OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Exercise 10 (B)

Solve the Following Equations:

Question 1. \( x^4 - 5x^2 + 6 = 0 \).
Answer: Given equation is \( x^4 - 5x^2 + 6 = 0 \) ... (1)
Let's substitute \( x^2 = t \) into equation (1). This gives us a new quadratic equation:
\( t^2 - 5t + 6 = 0 \)
Now, we factor this quadratic equation:
\( (t - 2)(t - 3) = 0 \)
This means either \( t - 2 = 0 \) or \( t - 3 = 0 \).
So, \( t = 2 \) or \( t = 3 \).
Since we defined \( t = x^2 \), we substitute back:
\( x^2 = 2 \) or \( x^2 = 3 \)
Taking the square root of both sides for each case:
\( x = \pm \sqrt{2} \) or \( x = \pm \sqrt{3} \)
Thus, the solutions for x are \( \pm \sqrt{2} \) and \( \pm \sqrt{3} \).
In simple words: We changed \( x^2 \) to a new letter, \( t \), to make the equation simpler to solve, like a regular quadratic equation. After finding values for \( t \), we put \( x^2 \) back in to find the final answers for \( x \).

🎯 Exam Tip: When dealing with equations involving higher powers like \( x^4 \), look for opportunities to make a substitution (like \( t = x^2 \)) to transform it into a standard quadratic equation. Remember to substitute back to find the values of the original variable.

Question 2. \( x^5 + 242 = \frac{243}{x^5} \).
Answer: Given equation is \( x^5 + 242 = \frac{243}{x^5} \) ... (1)
Let's substitute \( x^5 = y \) into equation (1):
\( y + 242 = \frac{243}{y} \)
Multiply both sides by \( y \) to remove the fraction:
\( y^2 + 242y = 243 \)
Rearrange the terms to form a quadratic equation:
\( y^2 + 242y - 243 = 0 \)
Now, we can factor this quadratic equation. We need two numbers that multiply to -243 and add up to 242. These numbers are 243 and -1.
\( y^2 - y + 243y - 243 = 0 \)
Factor by grouping:
\( y(y - 1) + 243(y - 1) = 0 \)
\( (y - 1)(y + 243) = 0 \)
This gives us two possible values for \( y \):
\( y - 1 = 0 \implies y = 1 \)
\( y + 243 = 0 \implies y = -243 \)
Since we know \( y = x^5 \), we substitute back these values:
Case 1: \( x^5 = 1 \)
\( \implies x = 1 \)
Case 2: \( x^5 = -243 \)
We know that \( (-3)^5 = -243 \).
\( \implies x^5 = (-3)^5 \implies x = -3 \)
Thus, the solutions for x are \( 1 \) and \( -3 \).
In simple words: We replaced \( x^5 \) with \( y \) to turn the problem into a simple quadratic equation. After solving for \( y \), we put \( x^5 \) back to find the actual values for \( x \).

🎯 Exam Tip: When dealing with fractional equations where the variable appears in the denominator, always multiply by the denominator to clear it. Be careful with signs when factoring quadratic equations, especially when the constant term is negative.

Question 3. \( 10x^{-2} - 9 - x^{-4} = 0 \).
Answer: Given equation is \( 10x^{-2} - 9 - x^{-4} = 0 \) ... (1)
This equation can be rewritten as \( -x^{-4} + 10x^{-2} - 9 = 0 \). Multiplying by -1, we get \( x^{-4} - 10x^{-2} + 9 = 0 \).
Let's substitute \( x^{-2} = t \) into equation (1). Then \( x^{-4} = (x^{-2})^2 = t^2 \).
So the equation becomes:
\( t^2 - 10t + 9 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to 9 and add up to -10. These numbers are -9 and -1.
\( t^2 - t - 9t + 9 = 0 \)
Factor by grouping:
\( t(t - 1) - 9(t - 1) = 0 \)
\( (t - 9)(t - 1) = 0 \)
This gives us two possible values for \( t \):
\( t - 9 = 0 \implies t = 9 \)
\( t - 1 = 0 \implies t = 1 \)
Since we defined \( t = x^{-2} \), we substitute back:
Case 1: \( x^{-2} = 9 \)
This means \( \frac{1}{x^2} = 9 \).
\( \implies x^2 = \frac{1}{9} \)
Taking the square root:
\( x = \pm \sqrt{\frac{1}{9}} \implies x = \pm \frac{1}{3} \)
Case 2: \( x^{-2} = 1 \)
This means \( \frac{1}{x^2} = 1 \).
\( \implies x^2 = 1 \)
Taking the square root:
\( x = \pm \sqrt{1} \implies x = \pm 1 \)
Thus, the solutions for x are \( \pm \frac{1}{3} \) and \( \pm 1 \).
In simple words: We changed terms with negative powers into a simple letter, \( t \), which helped us solve it like a standard quadratic equation. Then, we put the original powers back to find the answers for \( x \).

🎯 Exam Tip: Remember that \( x^{-n} = \frac{1}{x^n} \). This reciprocal relationship is key when dealing with negative exponents. Always write your final answers with both positive and negative roots for even powers.

Question 4. \( 3^{2x} - 10 \times 3^x + 9 = 0 \).
Answer: Given equation is \( 3^{2x} - 10 \times 3^x + 9 = 0 \) ... (1)
We can rewrite \( 3^{2x} \) as \( (3^x)^2 \).
Let's substitute \( 3^x = t \) into equation (1):
\( t^2 - 10t + 9 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to 9 and add up to -10. These numbers are -9 and -1.
\( t^2 - t - 9t + 9 = 0 \)
Factor by grouping:
\( t(t - 1) - 9(t - 1) = 0 \)
\( (t - 9)(t - 1) = 0 \)
This gives us two possible values for \( t \):
\( t - 9 = 0 \implies t = 9 \)
\( t - 1 = 0 \implies t = 1 \)
Since we defined \( t = 3^x \), we substitute back:
Case 1: \( 3^x = 9 \)
We know that \( 9 = 3^2 \).
\( \implies 3^x = 3^2 \)
Since the bases are the same, the exponents must be equal:
\( x = 2 \)
Case 2: \( 3^x = 1 \)
We know that any non-zero number raised to the power of 0 is 1. So \( 1 = 3^0 \).
\( \implies 3^x = 3^0 \)
Since the bases are the same, the exponents must be equal:
\( x = 0 \)
Thus, the solutions for x are \( 0 \) and \( 2 \).
In simple words: We saw that the equation had parts like \( 3^x \) and \( (3^x)^2 \). We replaced \( 3^x \) with \( t \) to make it a simple quadratic equation. After solving for \( t \), we put \( 3^x \) back and found the values for \( x \) using the rules of exponents.

🎯 Exam Tip: Equations involving exponents can often be simplified by substituting a new variable for the base raised to the power of x. Remember the rules of exponents, especially that \( a^0 = 1 \) for any non-zero \( a \), and \( a^{mx} = (a^m)^x \).

Question 5. \( 2^{2x-1} - 9 \times 2^{x-2} + 1 = 0 \).
Answer: Given equation is \( 2^{2x-1} - 9 \times 2^{x-2} + 1 = 0 \) ... (1)
We can rewrite the exponential terms using exponent rules:
\( 2^{2x-1} = 2^{2x} \times 2^{-1} = (2^x)^2 \times \frac{1}{2} \)
\( 2^{x-2} = 2^x \times 2^{-2} = 2^x \times \frac{1}{4} \)
Substitute these into the original equation:
\( (2^x)^2 \times \frac{1}{2} - 9 \times 2^x \times \frac{1}{4} + 1 = 0 \)
Let's substitute \( 2^x = t \):
\( t^2 \times \frac{1}{2} - 9 \times t \times \frac{1}{4} + 1 = 0 \)
\( \frac{t^2}{2} - \frac{9t}{4} + 1 = 0 \)
To clear the denominators, multiply the entire equation by 4:
\( 4 \times (\frac{t^2}{2}) - 4 \times (\frac{9t}{4}) + 4 \times 1 = 0 \times 4 \)
\( 2t^2 - 9t + 4 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to \( 2 \times 4 = 8 \) and add up to -9. These numbers are -8 and -1.
\( 2t^2 - 8t - t + 4 = 0 \)
Factor by grouping:
\( 2t(t - 4) - 1(t - 4) = 0 \)
\( (2t - 1)(t - 4) = 0 \)
This gives us two possible values for \( t \):
\( 2t - 1 = 0 \implies 2t = 1 \implies t = \frac{1}{2} \)
\( t - 4 = 0 \implies t = 4 \)
Since we defined \( t = 2^x \), we substitute back:
Case 1: \( 2^x = \frac{1}{2} \)
We know that \( \frac{1}{2} = 2^{-1} \).
\( \implies 2^x = 2^{-1} \)
Since the bases are the same, the exponents must be equal:
\( x = -1 \)
Case 2: \( 2^x = 4 \)
We know that \( 4 = 2^2 \).
\( \implies 2^x = 2^2 \)
Since the bases are the same, the exponents must be equal:
\( x = 2 \)
Thus, the solutions for x are \( -1 \) and \( 2 \).
In simple words: We used exponent rules to break down the tricky parts of the equation into simpler forms involving \( 2^x \). Then, we replaced \( 2^x \) with \( t \) to make it a quadratic equation, which we solved. Finally, we put \( 2^x \) back to find the actual \( x \) values.

🎯 Exam Tip: Always simplify exponential terms using properties like \( a^{m-n} = a^m \times a^{-n} \) and \( (a^m)^n = a^{mn} \) before making substitutions. Clearing denominators early can simplify factoring quadratic equations.

Question 6. \( 3^{2x+1} + 3^2 = 3^{x+3} + 3^x \).
Answer: Given equation is \( 3^{2x+1} + 3^2 = 3^{x+3} + 3^x \)
First, let's simplify the exponential terms using properties of exponents:
\( 3^{2x+1} = 3^{2x} \times 3^1 = (3^x)^2 \times 3 \)
\( 3^{x+3} = 3^x \times 3^3 \)
Substitute these into the equation:
\( (3^x)^2 \times 3 + 9 = 3^x \times 3^3 + 3^x \)
\( 3 \times (3^x)^2 + 9 = 3^x \times 27 + 3^x \)
Now, let's substitute \( 3^x = t \):
\( 3t^2 + 9 = 27t + t \)
\( 3t^2 + 9 = 28t \)
Rearrange the terms to form a quadratic equation:
\( 3t^2 - 28t + 9 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to \( 3 \times 9 = 27 \) and add up to -28. These numbers are -27 and -1.
\( 3t^2 - 27t - t + 9 = 0 \)
Factor by grouping:
\( 3t(t - 9) - 1(t - 9) = 0 \)
\( (t - 9)(3t - 1) = 0 \)
This gives us two possible values for \( t \):
\( t - 9 = 0 \implies t = 9 \)
\( 3t - 1 = 0 \implies 3t = 1 \implies t = \frac{1}{3} \)
Since we defined \( t = 3^x \), we substitute back:
Case 1: \( 3^x = 9 \)
We know that \( 9 = 3^2 \).
\( \implies 3^x = 3^2 \)
Since the bases are the same, the exponents must be equal:
\( x = 2 \)
Case 2: \( 3^x = \frac{1}{3} \)
We know that \( \frac{1}{3} = 3^{-1} \).
\( \implies 3^x = 3^{-1} \)
Since the bases are the same, the exponents must be equal:
\( x = -1 \)
Thus, the solutions for x are \( 2 \) and \( -1 \).
In simple words: First, we made the equation easier by separating the powers of 3. Then, we replaced \( 3^x \) with a simple letter, \( t \), to get a normal quadratic equation. After solving for \( t \), we put \( 3^x \) back to find the actual values of \( x \) using exponent rules.

🎯 Exam Tip: When simplifying exponential expressions, remember that \( a^{m+n} = a^m \times a^n \). Group terms involving the same exponential variable (e.g., \( 3^x \)) to create a quadratic form. Factoring by grouping is a useful technique for quadratic equations where the leading coefficient is not 1.

Question 7. \( \sqrt{x^2-3x} = 4x^2-12x-3 \).
Answer: Given equation is \( \sqrt{x^2-3x} = 4x^2-12x-3 \) ... (1)
Notice that \( 4x^2-12x = 4(x^2-3x) \). This suggests a substitution.
Let's substitute \( x^2-3x = t \) into equation (1).
The equation becomes:
\( \sqrt{t} = 4t - 3 \)
To remove the square root, we square both sides of the equation:
\( (\sqrt{t})^2 = (4t - 3)^2 \)
\( t = 16t^2 - 24t + 9 \)
Rearrange the terms to form a quadratic equation:
\( 16t^2 - 24t - t + 9 = 0 \)
\( 16t^2 - 25t + 9 = 0 \)
Now, we solve this quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a=16, b=-25, c=9 \).
\( t = \frac{-(-25) \pm \sqrt{(-25)^2 - 4 \times 16 \times 9}}{2 \times 16} \)
\( t = \frac{25 \pm \sqrt{625 - 576}}{32} \)
\( t = \frac{25 \pm \sqrt{49}}{32} \)
\( t = \frac{25 \pm 7}{32} \)
This gives us two possible values for \( t \):
Case 1: \( t = \frac{25 + 7}{32} = \frac{32}{32} = 1 \)
Case 2: \( t = \frac{25 - 7}{32} = \frac{18}{32} = \frac{9}{16} \)
Now, we substitute back \( t = x^2-3x \) for each case:

**Case I: When \( t = 1 \)**
\( x^2 - 3x = 1 \)
\( x^2 - 3x - 1 = 0 \)
Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a=1, b=-3, c=-1 \).
\( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times (-1)}}{2 \times 1} \)
\( x = \frac{3 \pm \sqrt{9 + 4}}{2} \)
\( x = \frac{3 \pm \sqrt{13}}{2} \)

**Case II: When \( t = \frac{9}{16} \)**
\( x^2 - 3x = \frac{9}{16} \)
Multiply by 16 to clear the fraction:
\( 16x^2 - 48x = 9 \)
\( 16x^2 - 48x - 9 = 0 \)
Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a=16, b=-48, c=-9 \).
\( x = \frac{-(-48) \pm \sqrt{(-48)^2 - 4 \times 16 \times (-9)}}{2 \times 16} \)
\( x = \frac{48 \pm \sqrt{2304 + 576}}{32} \)
\( x = \frac{48 \pm \sqrt{2880}}{32} \)
To simplify \( \sqrt{2880} \): \( \sqrt{2880} = \sqrt{64 \times 45} = \sqrt{64 \times 9 \times 5} = 8 \times 3 \sqrt{5} = 24\sqrt{5} \)
\( x = \frac{48 \pm 24\sqrt{5}}{32} \)
Divide the numerator and denominator by 8:
\( x = \frac{6 \pm 3\sqrt{5}}{4} \)
Thus, the solutions for x are \( \frac{3 \pm \sqrt{13}}{2} \) and \( \frac{6 \pm 3\sqrt{5}}{4} \).
It is important to check for extraneous solutions. Since we squared both sides, we must verify if the solutions for x satisfy the original equation, especially \( \sqrt{t} = 4t-3 \). For \( \sqrt{t} \) to be a real number, \( t \) must be non-negative. Both \( t=1 \) and \( t=9/16 \) are positive. Also, \( 4t-3 \) must be non-negative. For \( t=1 \), \( 4(1)-3 = 1 \ge 0 \). For \( t=9/16 \), \( 4(9/16)-3 = 9/4-3 = 9/4-12/4 = -3/4 < 0 \). This means \( t=9/16 \) is an extraneous solution for the equation \( \sqrt{t} = 4t-3 \). Therefore, the solutions derived from \( t=9/16 \) are not valid for the original equation. We only consider solutions from \( t=1 \).
The final valid solutions for x are \( \frac{3 \pm \sqrt{13}}{2} \).
In simple words: We noticed a repeated pattern in the equation and replaced that part with \( t \). After simplifying and squaring both sides, we solved a quadratic equation for \( t \). We had to put back the original expression to find \( x \). We also needed to check if any answers were "extra" because squaring can sometimes create solutions that don't work in the first equation.

🎯 Exam Tip: When an equation involves square roots, always isolate the square root term and then square both sides. This often leads to a quadratic equation. Crucially, always check your final solutions in the original equation to identify and reject any extraneous roots introduced by squaring, especially if you have terms like \( \sqrt{A} = B \) where B must be positive.

Question 8. \( \sqrt{\frac{x^2+2}{x^2-2}}+6 \sqrt{\frac{x^2-2}{x^2+2}} = 5 \).
Answer: Given equation is \( \sqrt{\frac{x^2+2}{x^2-2}}+6 \sqrt{\frac{x^2-2}{x^2+2}} = 5 \) ... (1)
Notice that the second term is the reciprocal of the first term inside the square root. Let's make a substitution.
Let \( y = \sqrt{\frac{x^2+2}{x^2-2}} \). Then \( \sqrt{\frac{x^2-2}{x^2+2}} = \frac{1}{y} \).
Substitute these into equation (1):
\( y + 6 \left(\frac{1}{y}\right) = 5 \)
\( y + \frac{6}{y} = 5 \)
Multiply the entire equation by \( y \) to clear the denominator (assuming \( y \neq 0 \)):
\( y^2 + 6 = 5y \)
Rearrange the terms to form a quadratic equation:
\( y^2 - 5y + 6 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3.
\( (y - 2)(y - 3) = 0 \)
This gives us two possible values for \( y \):
\( y - 2 = 0 \implies y = 2 \)
\( y - 3 = 0 \implies y = 3 \)
Now, we substitute back \( y = \sqrt{\frac{x^2+2}{x^2-2}} \) for each case:

**Case I: When \( y = 2 \)**
\( \sqrt{\frac{x^2+2}{x^2-2}} = 2 \)
Square both sides to remove the square root:
\( \left(\sqrt{\frac{x^2+2}{x^2-2}}\right)^2 = 2^2 \)
\( \frac{x^2+2}{x^2-2} = 4 \)
Multiply both sides by \( x^2-2 \):
\( x^2 + 2 = 4(x^2 - 2) \)
\( x^2 + 2 = 4x^2 - 8 \)
Group like terms:
\( 2 + 8 = 4x^2 - x^2 \)
\( 10 = 3x^2 \)
\( x^2 = \frac{10}{3} \)
Taking the square root:
\( x = \pm \sqrt{\frac{10}{3}} \)

**Case II: When \( y = 3 \)**
\( \sqrt{\frac{x^2+2}{x^2-2}} = 3 \)
Square both sides to remove the square root:
\( \left(\sqrt{\frac{x^2+2}{x^2-2}}\right)^2 = 3^2 \)
\( \frac{x^2+2}{x^2-2} = 9 \)
Multiply both sides by \( x^2-2 \):
\( x^2 + 2 = 9(x^2 - 2) \)
\( x^2 + 2 = 9x^2 - 18 \)
Group like terms:
\( 2 + 18 = 9x^2 - x^2 \)
\( 20 = 8x^2 \)
\( x^2 = \frac{20}{8} \)
Simplify the fraction:
\( x^2 = \frac{5}{2} \)
Taking the square root:
\( x = \pm \sqrt{\frac{5}{2}} \)
Thus, the solutions for x are \( \pm \sqrt{\frac{10}{3}} \) and \( \pm \sqrt{\frac{5}{2}} \).
In simple words: We noticed that part of the equation was the reverse of another part. We replaced this repeating part with a letter, \( y \), which made the equation a lot simpler to solve for \( y \). Then, we put the original expression back in place of \( y \) and squared both sides to find the values for \( x \).

🎯 Exam Tip: Recognize reciprocal relationships within complex expressions, especially under square roots. Substituting a variable for one part and its reciprocal for the other often simplifies the problem to a basic quadratic equation. Always remember to square both sides to eliminate square roots and solve for the original variable.

Question 9. \( \sqrt{\frac{2x^2+1}{x^2-1}}+6 \sqrt{\frac{x^2-1}{2x^2+1}} = 5 \).
Answer: Given equation is \( \sqrt{\frac{2x^2+1}{x^2-1}}+6 \sqrt{\frac{x^2-1}{2x^2+1}} = 5 \) ... (1)
Similar to the previous question, notice that the second term is the reciprocal of the first term inside the square root.
Let \( t = \sqrt{\frac{2x^2+1}{x^2-1}} \). Then \( \sqrt{\frac{x^2-1}{2x^2+1}} = \frac{1}{t} \).
Substitute these into equation (1):
\( t + 6 \left(\frac{1}{t}\right) = 5 \)
\( t + \frac{6}{t} = 5 \)
Multiply the entire equation by \( t \) to clear the denominator (assuming \( t \neq 0 \)):
\( t^2 + 6 = 5t \)
Rearrange the terms to form a quadratic equation:
\( t^2 - 5t + 6 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3.
\( (t - 2)(t - 3) = 0 \)
This gives us two possible values for \( t \):
\( t - 2 = 0 \implies t = 2 \)
\( t - 3 = 0 \implies t = 3 \)
Now, we substitute back \( t = \sqrt{\frac{2x^2+1}{x^2-1}} \) for each case:

**Case I: When \( t = 2 \)**
\( \sqrt{\frac{2x^2+1}{x^2-1}} = 2 \)
Square both sides to remove the square root:
\( \left(\sqrt{\frac{2x^2+1}{x^2-1}}\right)^2 = 2^2 \)
\( \frac{2x^2+1}{x^2-1} = 4 \)
Multiply both sides by \( x^2-1 \):
\( 2x^2 + 1 = 4(x^2 - 1) \)
\( 2x^2 + 1 = 4x^2 - 4 \)
Group like terms:
\( 1 + 4 = 4x^2 - 2x^2 \)
\( 5 = 2x^2 \)
\( x^2 = \frac{5}{2} \)
Taking the square root:
\( x = \pm \sqrt{\frac{5}{2}} \)

**Case II: When \( t = 3 \)**
\( \sqrt{\frac{2x^2+1}{x^2-1}} = 3 \)
Square both sides to remove the square root:
\( \left(\sqrt{\frac{2x^2+1}{x^2-1}}\right)^2 = 3^2 \)
\( \frac{2x^2+1}{x^2-1} = 9 \)
Multiply both sides by \( x^2-1 \):
\( 2x^2 + 1 = 9(x^2 - 1) \)
\( 2x^2 + 1 = 9x^2 - 9 \)
Group like terms:
\( 1 + 9 = 9x^2 - 2x^2 \)
\( 10 = 7x^2 \)
\( x^2 = \frac{10}{7} \)
Taking the square root:
\( x = \pm \sqrt{\frac{10}{7}} \)
Thus, the solutions for x are \( \pm \sqrt{\frac{5}{2}} \) and \( \pm \sqrt{\frac{10}{7}} \).
In simple words: This problem was solved by noticing the repeating pattern, where one part of the equation was the inverse of another. We used a temporary letter \( t \) to simplify it into a quadratic equation. After finding \( t \), we put the original expression back and squared both sides to get the final \( x \) values.

🎯 Exam Tip: Pay close attention to patterns in equations, especially those with square roots and fractions. Recognizing a term and its reciprocal allows for a quick substitution to convert complex equations into simpler quadratic forms. Always isolate the variable and simplify fractions at the end.

Question 10. \( x (x - 1) (x + 2) (x - 3) + 8 = 0 \).
Answer: Given equation is \( x (x - 1) (x + 2) (x - 3) + 8 = 0 \).
We need to group the terms in a way that creates a common expression. Let's rearrange the factors and multiply them:
\( [x (x + 2)] \times [(x - 1) (x - 3)] + 8 = 0 \)
\( [x^2 + 2x] \times [x^2 - 3x - x + 3] + 8 = 0 \)
\( [x^2 + 2x] \times [x^2 - 4x + 3] + 8 = 0 \)
This doesn't seem to create a simple common term. Let's try another grouping:
Group factors with sums of constants that are equal:
\( x \) and \( (x - 1) \), \( (x + 2) \) and \( (x - 3) \) -> \( 0-1 = -1 \), \( 2-3 = -1 \). This pairing works well.
\( [x (x - 1)] \times [(x + 2) (x - 3)] + 8 = 0 \)
\( [x^2 - x] \times [x^2 - 3x + 2x - 6] + 8 = 0 \)
\( [x^2 - x] \times [x^2 - x - 6] + 8 = 0 \) ... (1)
Now, we have a common expression \( x^2 - x \). Let's substitute \( x^2 - x = t \) into equation (1).
The equation becomes:
\( t(t - 6) + 8 = 0 \)
\( t^2 - 6t + 8 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to 8 and add up to -6. These numbers are -4 and -2.
\( t^2 - 4t - 2t + 8 = 0 \)
Factor by grouping:
\( t(t - 4) - 2(t - 4) = 0 \)
\( (t - 4)(t - 2) = 0 \)
This gives us two possible values for \( t \):
\( t - 4 = 0 \implies t = 4 \)
\( t - 2 = 0 \implies t = 2 \)
Now, we substitute back \( t = x^2 - x \) for each case:

**Case I: When \( t = 4 \)**
\( x^2 - x = 4 \)
\( x^2 - x - 4 = 0 \)
Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a=1, b=-1, c=-4 \).
\( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times (-4)}}{2 \times 1} \)
\( x = \frac{1 \pm \sqrt{1 + 16}}{2} \)
\( x = \frac{1 \pm \sqrt{17}}{2} \)

**Case II: When \( t = 2 \)**
\( x^2 - x = 2 \)
\( x^2 - x - 2 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to -2 and add up to -1. These numbers are -2 and 1.
\( (x - 2)(x + 1) = 0 \)
This gives us two possible values for \( x \):
\( x - 2 = 0 \implies x = 2 \)
\( x + 1 = 0 \implies x = -1 \)
Thus, the solutions for x are \( -1, 2, \frac{1 \pm \sqrt{17}}{2} \).
In simple words: We grouped the terms cleverly so that we got a repeating part like \( x^2 - x \). We replaced this with \( t \) to solve a simpler quadratic equation. After finding \( t \), we put \( x^2 - x \) back and solved for \( x \) using either factoring or the quadratic formula.

🎯 Exam Tip: For equations involving products of four linear factors, try grouping pairs of factors such that the sum of the constant terms in each pair is equal. This often leads to a common quadratic expression, simplifying the problem into a substitution and a standard quadratic solution.

Question 11. \( (x - 7) (x - 3) (x + 1) (x + 5) = 1680 \).
Answer: Given equation is \( (x - 7) (x - 3) (x + 1) (x + 5) = 1680 \).
Similar to the previous problem, we group the factors such that the sum of the constants in each pair is the same. Let's try pairing \( (x-7) \) with \( (x+5) \) (sum of constants is \( -7+5 = -2 \)) and \( (x-3) \) with \( (x+1) \) (sum of constants is \( -3+1 = -2 \)).
\( [(x - 7)(x + 5)] \times [(x - 3)(x + 1)] = 1680 \)
Multiply the pairs:
\( [x^2 + 5x - 7x - 35] \times [x^2 + x - 3x - 3] = 1680 \)
\( [x^2 - 2x - 35] \times [x^2 - 2x - 3] = 1680 \) ... (1)
Now, we have a common expression \( x^2 - 2x \). Let's substitute \( x^2 - 2x = t \) into equation (1).
The equation becomes:
\( (t - 35)(t - 3) = 1680 \)
Expand the left side:
\( t^2 - 3t - 35t + 105 = 1680 \)
\( t^2 - 38t + 105 = 1680 \)
Rearrange to form a quadratic equation:
\( t^2 - 38t + 105 - 1680 = 0 \)
\( t^2 - 38t - 1575 = 0 \)
Now, we solve this quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a=1, b=-38, c=-1575 \).
\( t = \frac{-(-38) \pm \sqrt{(-38)^2 - 4 \times 1 \times (-1575)}}{2 \times 1} \)
\( t = \frac{38 \pm \sqrt{1444 + 6300}}{2} \)
\( t = \frac{38 \pm \sqrt{7744}}{2} \)
The square root of 7744 is 88.
\( t = \frac{38 \pm 88}{2} \)
This gives us two possible values for \( t \):
Case 1: \( t = \frac{38 + 88}{2} = \frac{126}{2} = 63 \)
Case 2: \( t = \frac{38 - 88}{2} = \frac{-50}{2} = -25 \)
Now, we substitute back \( t = x^2 - 2x \) for each case:

**Case I: When \( t = 63 \)**
\( x^2 - 2x = 63 \)
\( x^2 - 2x - 63 = 0 \)
Factor this quadratic equation. We need two numbers that multiply to -63 and add up to -2. These numbers are -9 and 7.
\( (x - 9)(x + 7) = 0 \)
This gives us two possible values for \( x \):
\( x - 9 = 0 \implies x = 9 \)
\( x + 7 = 0 \implies x = -7 \)

**Case II: When \( t = -25 \)**
\( x^2 - 2x = -25 \)
\( x^2 - 2x + 25 = 0 \)
Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a=1, b=-2, c=25 \).
\( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 1 \times 25}}{2 \times 1} \)
\( x = \frac{2 \pm \sqrt{4 - 100}}{2} \)
\( x = \frac{2 \pm \sqrt{-96}}{2} \)
Since the discriminant (\( \sqrt{-96} \)) is negative, there are no real solutions in this case. We can express the complex solutions:
\( \sqrt{-96} = \sqrt{16 \times 6 \times -1} = 4\sqrt{6}i \)
\( x = \frac{2 \pm 4\sqrt{6}i}{2} \)
\( x = 1 \pm 2\sqrt{6}i \)
Thus, the real solutions for x are \( 9 \) and \( -7 \). The complex solutions are \( 1 \pm 2\sqrt{6}i \).
In simple words: We organized the numbers in the equation to find a common part, \( x^2 - 2x \), which we called \( t \). This made a simpler equation that we solved for \( t \). After putting back \( x^2 - 2x \), we solved for \( x \). For some values of \( t \), we found real answers for \( x \), but for others, the answers were complex numbers (not real).

🎯 Exam Tip: When grouping factors in such products, ensure the sums of the constant terms in each pair are equal. This strategy consistently yields a common quadratic expression for substitution. Be prepared to handle both real and complex solutions, as quadratic equations can have either. The discriminant \( b^2-4ac \) tells you the nature of the roots (positive for real, negative for complex).

Question 12. \( (2x - 7) (x^2 - 9) (2x + 5) = 91 \).
Answer: Given equation is \( (2x - 7) (x^2 - 9) (2x + 5) = 91 \).
First, factor \( x^2 - 9 \) as a difference of squares: \( (x-3)(x+3) \).
The equation becomes: \( (2x - 7) (x - 3) (x + 3) (2x + 5) = 91 \).
Now, we group the factors such that the sum of the products of the coefficients of x and the constants are similar. Let's try pairing \( (2x-7) \) with \( (x+3) \) and \( (x-3) \) with \( (2x+5) \).
\( [(2x - 7)(x + 3)] \times [(x - 3)(2x + 5)] = 91 \)
Multiply the pairs:
\( [2x^2 + 6x - 7x - 21] \times [2x^2 + 5x - 6x - 15] = 91 \)
\( [2x^2 - x - 21] \times [2x^2 - x - 15] = 91 \) ... (1)
Now, we have a common expression \( 2x^2 - x \). Let's substitute \( 2x^2 - x = t \) into equation (1).
The equation becomes:
\( (t - 21)(t - 15) = 91 \)
Expand the left side:
\( t^2 - 15t - 21t + 315 = 91 \)
\( t^2 - 36t + 315 = 91 \)
Rearrange to form a quadratic equation:
\( t^2 - 36t + 315 - 91 = 0 \)
\( t^2 - 36t + 224 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to 224 and add up to -36. These numbers are -28 and -8.
\( t^2 - 28t - 8t + 224 = 0 \)
Factor by grouping:
\( t(t - 28) - 8(t - 28) = 0 \)
\( (t - 28)(t - 8) = 0 \)
This gives us two possible values for \( t \):
\( t - 28 = 0 \implies t = 28 \)
\( t - 8 = 0 \implies t = 8 \)
Now, we substitute back \( t = 2x^2 - x \) for each case:

**Case I: When \( t = 8 \)**
\( 2x^2 - x = 8 \)
\( 2x^2 - x - 8 = 0 \)
Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a=2, b=-1, c=-8 \).
\( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 2 \times (-8)}}{2 \times 2} \)
\( x = \frac{1 \pm \sqrt{1 + 64}}{4} \)
\( x = \frac{1 \pm \sqrt{65}}{4} \)

**Case II: When \( t = 28 \)**
\( 2x^2 - x = 28 \)
\( 2x^2 - x - 28 = 0 \)
Factor this quadratic equation. We need two numbers that multiply to \( 2 \times -28 = -56 \) and add up to -1. These numbers are -8 and 7.
\( 2x^2 - 8x + 7x - 28 = 0 \)
Factor by grouping:
\( 2x(x - 4) + 7(x - 4) = 0 \)
\( (x - 4)(2x + 7) = 0 \)
This gives us two possible values for \( x \):
\( x - 4 = 0 \implies x = 4 \)
\( 2x + 7 = 0 \implies 2x = -7 \implies x = -\frac{7}{2} \)
Thus, the solutions for x are \( 4, -\frac{7}{2}, \frac{1 \pm \sqrt{65}}{4} \).
In simple words: We first broke down one part of the equation into two factors. Then, we cleverly matched the factors to create a common expression, \( 2x^2 - x \). We called this \( t \) to make the equation simpler and solved for \( t \). Finally, we put the original expression back and found the values for \( x \).

🎯 Exam Tip: Always look for ways to factor expressions like \( x^2-9 \) first. When grouping multiple linear factors, strategically pair them to produce a common quadratic term. This common term then becomes a suitable candidate for substitution, simplifying the entire problem into a solvable quadratic equation.

Question 13. By substituting \( y = 2^x \), or otherwise, solve the equation \( 2^{2x} + 2^{x+2} - 4 \times 2^3 = 0 \).
Answer: Given equation is \( 2^{2x} + 2^{x+2} - 4 \times 2^3 = 0 \) ... (1)
First, let's simplify the exponential terms using properties of exponents:
\( 2^{2x} = (2^x)^2 \)
\( 2^{x+2} = 2^x \times 2^2 \)
\( 4 \times 2^3 = 2^2 \times 2^3 = 2^{2+3} = 2^5 = 32 \)
Substitute these into the equation:
\( (2^x)^2 + 2^x \times 2^2 - 32 = 0 \)
\( (2^x)^2 + 4 \times 2^x - 32 = 0 \)
Now, we are given to substitute \( y = 2^x \). So, the equation becomes:
\( y^2 + 4y - 32 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to -32 and add up to 4. These numbers are 8 and -4.
\( y^2 + 8y - 4y - 32 = 0 \)
Factor by grouping:
\( y(y + 8) - 4(y + 8) = 0 \)
\( (y + 8)(y - 4) = 0 \)
This gives us two possible values for \( y \):
\( y + 8 = 0 \implies y = -8 \)
\( y - 4 = 0 \implies y = 4 \)
Since we defined \( y = 2^x \), we substitute back:
Case 1: \( 2^x = -8 \)
An exponential term \( 2^x \) is always positive for any real value of \( x \). Therefore, \( 2^x = -8 \) has no real solution.
Case 2: \( 2^x = 4 \)
We know that \( 4 = 2^2 \).
\( \implies 2^x = 2^2 \)
Since the bases are the same, the exponents must be equal:
\( x = 2 \)
Thus, the only real solution for x is \( 2 \).
In simple words: We simplified the given equation using rules of exponents. Then, following the hint, we replaced \( 2^x \) with \( y \) to solve a simple quadratic equation. When we got back to \( 2^x = y \), we found that one answer for \( y \) didn't work because \( 2^x \) can never be a negative number, leaving us with only one real solution for \( x \).

🎯 Exam Tip: When solving exponential equations by substitution, always remember that an exponential expression like \( a^x \) (where \( a > 0 \)) must always be positive. If you obtain a negative value for your substituted variable, it means there are no real solutions for that particular case. Carefully simplify exponents before substitution.

Question 14. \( 2x^2 : 2^x = 8 : 1 \).
Answer: Given ratio is \( 2x^2 : 2^x = 8 : 1 \).
We can write the ratio as a fraction:
\( \frac{2^{x^2}}{2^x} = \frac{8}{1} \)
Using the exponent rule \( \frac{a^m}{a^n} = a^{m-n} \), the left side becomes:
\( 2^{x^2 - x} = 8 \)
We know that \( 8 = 2^3 \). Substitute this into the equation:
\( 2^{x^2 - x} = 2^3 \)
Since the bases are the same, the exponents must be equal:
\( x^2 - x = 3 \)
Rearrange the terms to form a quadratic equation:
\( x^2 - x - 3 = 0 \)
Now, we solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a=1, b=-1, c=-3 \).
\( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times (-3)}}{2 \times 1} \)
\( x = \frac{1 \pm \sqrt{1 + 12}}{2} \)
\( x = \frac{1 \pm \sqrt{13}}{2} \)
Thus, the solutions for x are \( \frac{1 \pm \sqrt{13}}{2} \).
In simple words: We first wrote the ratio as a fraction. Then, we used exponent rules to simplify the left side and wrote 8 as a power of 2. By matching the powers, we got a simple quadratic equation which we solved using the quadratic formula to find the values of \( x \).

🎯 Exam Tip: Remember that a ratio \( a:b \) can always be expressed as a fraction \( \frac{a}{b} \). For exponential terms, utilize exponent rules like \( \frac{a^m}{a^n} = a^{m-n} \) to simplify expressions. If the bases are equal, then their exponents must also be equal, which often leads to a simpler equation.

Question 15. \( 2^{2x+3} + 2^{x+3} = 1 + 2^x \).
Answer: Given equation is \( 2^{2x+3} + 2^{x+3} = 1 + 2^x \) ... (1)
First, let's simplify the exponential terms using properties of exponents:
\( 2^{2x+3} = 2^{2x} \times 2^3 = (2^x)^2 \times 8 \)
\( 2^{x+3} = 2^x \times 2^3 = 2^x \times 8 \)
Substitute these into the equation:
\( 8 \times (2^x)^2 + 8 \times 2^x = 1 + 2^x \)
Now, let's substitute \( 2^x = t \):
\( 8t^2 + 8t = 1 + t \)
Rearrange the terms to form a quadratic equation:
\( 8t^2 + 8t - t - 1 = 0 \)
\( 8t^2 + 7t - 1 = 0 \)
Now, we factor this quadratic equation. We need two numbers that multiply to \( 8 \times -1 = -8 \) and add up to 7. These numbers are 8 and -1.
\( 8t^2 + 8t - t - 1 = 0 \)
Factor by grouping:
\( 8t(t + 1) - 1(t + 1) = 0 \)
\( (t + 1)(8t - 1) = 0 \)
This gives us two possible values for \( t \):
\( t + 1 = 0 \implies t = -1 \)
\( 8t - 1 = 0 \implies 8t = 1 \implies t = \frac{1}{8} \)
Since we defined \( t = 2^x \), we substitute back:
Case 1: \( 2^x = -1 \)
An exponential term \( 2^x \) is always positive for any real value of \( x \). Therefore, \( 2^x = -1 \) has no real solution.
Case 2: \( 2^x = \frac{1}{8} \)
We know that \( \frac{1}{8} = \frac{1}{2^3} = 2^{-3} \).
\( \implies 2^x = 2^{-3} \)
Since the bases are the same, the exponents must be equal:
\( x = -3 \)
Thus, the only real solution for x is \( -3 \).
In simple words: We used exponent rules to simplify the equation's parts. Then, we replaced \( 2^x \) with \( t \) to solve a simple quadratic equation. When putting \( 2^x \) back, we found that one answer for \( t \) was impossible because \( 2^x \) cannot be negative, leaving us with only one real solution for \( x \).

🎯 Exam Tip: Always begin by breaking down exponential terms using the rules of exponents, such as \( a^{m+n} = a^m \times a^n \). This will help you identify a common base to substitute (e.g., \( 2^x \)). Remember to reject any solutions where an exponential term \( a^x \) (for \( a>0 \)) equates to a negative number, as this is not possible in real numbers.

Question 16. \( 4^x-3^{x-\frac{1}{2}} = 3^{x+\frac{1}{2}}-2^{2x-1} \).
Answer: Given equation is \( 4^x-3^{x-\frac{1}{2}} = 3^{x+\frac{1}{2}}-2^{2x-1} \).
First, let's simplify and rearrange the terms. Group terms with base 4 and base 2 on one side, and terms with base 3 on the other.
\( 4^x + 2^{2x-1} = 3^{x+\frac{1}{2}} + 3^{x-\frac{1}{2}} \)
Simplify each term using exponent rules:
\( 4^x = (2^2)^x = 2^{2x} \)
\( 2^{2x-1} = 2^{2x} \times 2^{-1} = 2^{2x} \times \frac{1}{2} \)
\( 3^{x+\frac{1}{2}} = 3^x \times 3^{\frac{1}{2}} = 3^x \sqrt{3} \)
\( 3^{x-\frac{1}{2}} = 3^x \times 3^{-\frac{1}{2}} = 3^x \times \frac{1}{\sqrt{3}} \)
Substitute these simplified terms back into the equation:
\( 2^{2x} + 2^{2x} \times \frac{1}{2} = 3^x \sqrt{3} + 3^x \frac{1}{\sqrt{3}} \)
Factor out common terms on both sides:
\( 2^{2x} \left(1 + \frac{1}{2}\right) = 3^x \left(\sqrt{3} + \frac{1}{\sqrt{3}}\right) \)
Simplify the terms in the parentheses:
\( 2^{2x} \left(\frac{2+1}{2}\right) = 3^x \left(\frac{\sqrt{3} \times \sqrt{3} + 1}{\sqrt{3}}\right) \)
\( 2^{2x} \left(\frac{3}{2}\right) = 3^x \left(\frac{3 + 1}{\sqrt{3}}\right) \)
\( 2^{2x} \left(\frac{3}{2}\right) = 3^x \left(\frac{4}{\sqrt{3}}\right) \)
Now, gather terms with \( 2^{2x} \) and \( 3^x \):
\( \frac{2^{2x}}{3^x} = \frac{4/\sqrt{3}}{3/2} \)
\( \frac{(2^2)^x}{3^x} = \frac{4}{\sqrt{3}} \times \frac{2}{3} \)
\( \frac{4^x}{3^x} = \frac{8}{3\sqrt{3}} \)
\( \left(\frac{4}{3}\right)^x = \frac{8}{3^{\frac{3}{2}}} \)
Rewrite the right side using powers of 2 and 3:
\( \left(\frac{4}{3}\right)^x = \frac{2^3}{3^{\frac{3}{2}}} = \frac{2^3}{(\sqrt{3})^3} = \left(\frac{2}{\sqrt{3}}\right)^3 \)
We also know that \( 4 = 2^2 \) and \( 3 = (\sqrt{3})^2 \).
So, \( \left(\frac{4}{3}\right)^x = \left(\frac{2^2}{(\sqrt{3})^2}\right)^x = \left(\left(\frac{2}{\sqrt{3}}\right)^2\right)^x = \left(\frac{2}{\sqrt{3}}\right)^{2x} \)
Therefore, the equation is:
\( \left(\frac{2}{\sqrt{3}}\right)^{2x} = \left(\frac{2}{\sqrt{3}}\right)^3 \)
Since the bases are the same, the exponents must be equal:
\( 2x = 3 \)
\( x = \frac{3}{2} \)
Thus, the solution for x is \( \frac{3}{2} \).
In simple words: We first grouped terms with similar bases and simplified each part using rules of exponents. We then factored out the common terms. After rearranging the equation, we matched the bases on both sides. Once the bases were the same, we set the powers equal to each other to find the value of \( x \).

🎯 Exam Tip: When dealing with equations involving multiple bases and fractional exponents, the key is to isolate terms by base (e.g., all terms with base 2 or 4 on one side, and base 3 terms on the other). Convert all bases to their simplest forms (e.g., 4 to \( 2^2 \)). Simplify sums of fractional exponents carefully. The goal is to reach a form \( a^x = a^k \), from which you can equate the exponents \( x = k \).

Solve the Following Equations:

Question 1. \(x^2 - 5x^2 + 6 = 0.\)
Answer: The given equation is \(x^4 - 5x^2 + 6 = 0\). We can make this simpler by letting \(x^2 = t\).
So, the equation becomes \(t^2 - 5t + 6 = 0\).
We can factorize this quadratic equation:
\(t^2 - 2t - 3t + 6 = 0\)
\(t(t - 2) - 3(t - 2) = 0\)
\((t - 2)(t - 3) = 0\)
This means either \(t - 2 = 0\) or \(t - 3 = 0\).
So, \(t = 2\) or \(t = 3\).
Now, we put \(x^2\) back instead of \(t\):
\(x^2 = 2\) or \(x^2 = 3\)
Taking the square root of both sides:
\(x = \pm \sqrt{2}\) or \(x = \pm \sqrt{3}\)
These are the solutions to the equation. Finding the roots by substitution helps simplify complex equations.
In simple words: First, we change \(x^2\) to \(t\) to make the equation easier. Then we solve for \(t\). After finding \(t\), we change it back to \(x^2\) and find the final values for \(x\).

🎯 Exam Tip: When an equation has terms like \(x^4\) and \(x^2\), consider using a substitution like \(t = x^2\) to transform it into a standard quadratic equation, which is easier to solve.

Question 2. \(x^5 + 242 = \frac{243}{x^5}.\)
Answer: The given equation is \(x^5 + 242 = \frac{243}{x^5}\).
Let's simplify this by setting \(x^5 = y\).
The equation becomes \(y + 242 = \frac{243}{y}\).
Now, multiply both sides by \(y\) to clear the fraction:
\(y^2 + 242y = 243\)
Bring all terms to one side to form a quadratic equation:
\(y^2 + 242y - 243 = 0\)
We can factorize this quadratic equation. We need two numbers that multiply to -243 and add up to 242. These are 243 and -1.
\(y^2 + 243y - y - 243 = 0\)
\(y(y + 243) - 1(y + 243) = 0\)
\((y - 1)(y + 243) = 0\)
This means either \(y - 1 = 0\) or \(y + 243 = 0\).
So, \(y = 1\) or \(y = -243\).
Now, substitute \(x^5\) back in place of \(y\):
\(x^5 = 1\) or \(x^5 = -243\)
We know that \(1 = 1^5\) and \(-243 = (-3)^5\).
So, \(x^5 = 1^5\) or \(x^5 = (-3)^5\)
This gives us the solutions for \(x\):
\(x = 1\) or \(x = -3\)
These are the real solutions for the equation. Sometimes, a variable appears with a specific power in different parts of an equation, making substitution very useful.
In simple words: We replace \(x^5\) with \(y\) to make the equation simpler, like a normal quadratic. We solve for \(y\), then put \(x^5\) back to find the final value of \(x\).

🎯 Exam Tip: Look for repeated terms with the same power in equations. Substituting a new variable for such a term often simplifies the equation into a solvable quadratic form.

Question 3. \(10x^{-2} - 9 - x^4 = 0.\)
Answer: The given equation is \(10x^{-2} - 9 - x^4 = 0\).
Let's rearrange the terms in decreasing powers of x, and also rewrite \(x^{-2}\) as \(\frac{1}{x^2}\):
\(-x^4 - 9 + \frac{10}{x^2} = 0\)
Multiply by \(-1\) to make the leading term positive:
\(x^4 + 9 - \frac{10}{x^2} = 0\)
To solve this, let's substitute \(x^{-2} = t\). Then \(x^2 = \frac{1}{t}\) and \(x^4 = \frac{1}{t^2}\).
Substituting these into the equation:
\(\frac{1}{t^2} - 9 + 10t = 0\)
Multiply the entire equation by \(t^2\) (assuming \(t \neq 0\)):
\(1 - 9t^2 + 10t^3 = 0\)
Rearranging into a standard cubic form:
\(10t^3 - 9t^2 + 1 = 0\)
This looks like a cubic equation, but the original solution follows a different path. Let's re-examine the given equation and the provided solution's approach. The solution uses \(x^{-2} = t\). This means the original equation can be seen as: \(10x^{-2} - 9 - x^4 = 0\) There seems to be an OCR issue in the question as it appears in the PDF: `10x-2 – 9 – x^4 = 0`. The solution on page 3 clearly processes `10t - 9 - t^2 = 0`. This implies the original equation was `10x^{-2} - 9 - (x^{-2})^2 = 0`, or `10x^{-2} - 9 - x^{-4} = 0`. Given the solution's steps, it correctly processes the equation as if it was \(10x^{-2} - 9 - x^{-4} = 0\). I will follow the solution's logic for the answer. Let the given equation be \(10x^{-2} - 9 - x^{-4} = 0\). (Following the solution's implied question)
We can substitute \(x^{-2} = t\). Then \(x^{-4} = (x^{-2})^2 = t^2\).
The equation becomes \(10t - 9 - t^2 = 0\).
Rearranging it into a standard quadratic form:
\(-t^2 + 10t - 9 = 0\)
Multiply by \(-1\) to make the leading coefficient positive:
\(t^2 - 10t + 9 = 0\)
Factorize the quadratic equation:
\(t^2 - t - 9t + 9 = 0\)
\(t(t - 1) - 9(t - 1) = 0\)
\((t - 9)(t - 1) = 0\)
This means either \(t - 9 = 0\) or \(t - 1 = 0\).
So, \(t = 9\) or \(t = 1\).
Now, substitute \(x^{-2}\) back for \(t\):
\(x^{-2} = 9\) or \(x^{-2} = 1\)
This can be written as:
\(\frac{1}{x^2} = 9\) or \(\frac{1}{x^2} = 1\)
Solving for \(x^2\):
\(x^2 = \frac{1}{9}\) or \(x^2 = 1\)
Taking the square root of both sides:
\(x = \pm \sqrt{\frac{1}{9}}\) or \(x = \pm \sqrt{1}\)
\(x = \pm \frac{1}{3}\) or \(x = \pm 1\)
Thus, the solutions are \(x = \pm \frac{1}{3}, \pm 1\). Exponential equations can often be simplified with a clever substitution.
In simple words: We replace \(x^{-2}\) with \(t\) to get a simple quadratic equation. We solve this for \(t\). Then we put \(x^{-2}\) back to find the values of \(x\).

🎯 Exam Tip: When dealing with negative powers like \(x^{-2}\), remember that \(x^{-2} = \frac{1}{x^2}\). If you see terms like \(x^{-2}\) and \(x^{-4}\) (which is \((x^{-2})^2\)), use substitution to simplify.

Question 4. \(3^{2x} - 10 \times 3^x + 9 = 0.\)
Answer: The given equation is \(3^{2x} - 10 \times 3^x + 9 = 0\).
We can rewrite \(3^{2x}\) as \((3^x)^2\).
So the equation is \((3^x)^2 - 10 \times 3^x + 9 = 0\).
Let's substitute \(3^x = t\).
The equation becomes \(t^2 - 10t + 9 = 0\).
Now, we factorize this quadratic equation:
\(t^2 - t - 9t + 9 = 0\)
\(t(t - 1) - 9(t - 1) = 0\)
\((t - 9)(t - 1) = 0\)
This means either \(t - 9 = 0\) or \(t - 1 = 0\).
So, \(t = 9\) or \(t = 1\).
Now, substitute \(3^x\) back for \(t\):
\(3^x = 9\) or \(3^x = 1\)
We know that \(9 = 3^2\) and \(1 = 3^0\).
So, \(3^x = 3^2\) or \(3^x = 3^0\).
Comparing the powers, we get the values for \(x\):
\(x = 2\) or \(x = 0\)
Thus, the solutions are \(x = 0, 2\). This technique is often used to solve exponential equations.
In simple words: We let \(3^x\) be \(t\) to change the hard exponential equation into an easy quadratic one. We solve for \(t\), then change it back to \(3^x\) to find the values of \(x\).

🎯 Exam Tip: For exponential equations where the base and variable appear in different powers (e.g., \(a^{2x}\) and \(a^x\)), substitution of \(t=a^x\) is a common and effective strategy.

Question 5. \(2^{2x-1} - 9 \times 2^{x-2} + 1 = 0.\)
Answer: The given equation is \(2^{2x-1} - 9 \times 2^{x-2} + 1 = 0\).
We can use the exponent rule \(a^{m-n} = \frac{a^m}{a^n}\) and \(a^{m+n} = a^m \times a^n\).
Rewrite the terms:
\(2^{2x-1} = 2^{2x} \times 2^{-1} = (2^x)^2 \times \frac{1}{2}\)
\(2^{x-2} = 2^x \times 2^{-2} = 2^x \times \frac{1}{4}\)
Substitute these into the equation:
\((2^x)^2 \times \frac{1}{2} - 9 \times 2^x \times \frac{1}{4} + 1 = 0\)
Let's substitute \(2^x = t\).
The equation becomes \(t^2 \times \frac{1}{2} - 9 \times t \times \frac{1}{4} + 1 = 0\)
\(\frac{t^2}{2} - \frac{9t}{4} + 1 = 0\)
To remove the fractions, multiply the entire equation by the least common multiple of the denominators (2 and 4), which is 4:
\(4 \times \frac{t^2}{2} - 4 \times \frac{9t}{4} + 4 \times 1 = 0\)
\(2t^2 - 9t + 4 = 0\)
Now, we factorize this quadratic equation:
\(2t^2 - 8t - t + 4 = 0\)
\(2t(t - 4) - 1(t - 4) = 0\)
\((2t - 1)(t - 4) = 0\)
This means either \(2t - 1 = 0\) or \(t - 4 = 0\).
So, \(2t = 1 \implies t = \frac{1}{2}\)
or \(t = 4\).
Now, substitute \(2^x\) back for \(t\):
\(2^x = \frac{1}{2}\) or \(2^x = 4\)
We know that \(\frac{1}{2} = 2^{-1}\) and \(4 = 2^2\).
So, \(2^x = 2^{-1}\) or \(2^x = 2^2\).
Comparing the powers, we get the values for \(x\):
\(x = -1\) or \(x = 2\)
Therefore, the solutions are \(x = -1\) or \(x = 2\). Understanding exponent rules is key to solving these types of problems.
In simple words: First, we use rules of exponents to rewrite the terms. Then we let \(2^x\) be \(t\), which turns the equation into a simple quadratic. We solve for \(t\) and then change it back to \(2^x\) to find \(x\).

🎯 Exam Tip: Always simplify exponential terms using exponent rules like \(a^{m-n} = a^m/a^n\) before attempting substitution. This makes the quadratic form clearer and easier to manage.

Question 6. \(3^{2x+1} + 3^2 = 3^{x+3} + 3^x.\)
Answer: The given equation is \(3^{2x+1} + 3^2 = 3^{x+3} + 3^x\).
We can use the exponent rule \(a^{m+n} = a^m \times a^n\).
Rewrite the terms:
\(3^{2x+1} = 3^{2x} \times 3^1 = (3^x)^2 \times 3\)
\(3^{x+3} = 3^x \times 3^3 = 3^x \times 27\)
The equation becomes:
\(3 \times (3^x)^2 + 9 = 27 \times 3^x + 3^x\)
Let's substitute \(3^x = t\).
The equation becomes \(3t^2 + 9 = 27t + t\)
\(3t^2 + 9 = 28t\)
Rearrange into a standard quadratic form:
\(3t^2 - 28t + 9 = 0\)
Now, we factorize this quadratic equation:
\(3t^2 - 27t - t + 9 = 0\)
\(3t(t - 9) - 1(t - 9) = 0\)
\((t - 9)(3t - 1) = 0\)
This means either \(t - 9 = 0\) or \(3t - 1 = 0\).
So, \(t = 9\) or \(3t = 1 \implies t = \frac{1}{3}\).
Now, substitute \(3^x\) back for \(t\):
\(3^x = 9\) or \(3^x = \frac{1}{3}\)
We know that \(9 = 3^2\) and \(\frac{1}{3} = 3^{-1}\).
So, \(3^x = 3^2\) or \(3^x = 3^{-1}\).
Comparing the powers, we get the values for \(x\):
\(x = 2\) or \(x = -1\)
Thus, the solutions are \(x = 2, -1\). Grouping terms effectively helps in factorizing quadratic equations.
In simple words: We simplify the terms using rules of powers. Then we replace \(3^x\) with \(t\). This gives us a quadratic equation, which we solve for \(t\). Finally, we put \(3^x\) back to find the values for \(x\).

🎯 Exam Tip: Remember to gather all terms involving the substituted variable to one side to form a standard quadratic equation before attempting factorization.

Question 7. \(\sqrt{x^2-3x}=4x^2-12x-3-3.\)
Answer: The given equation is \(\sqrt{x^2-3x}=4x^2-12x-3-3\).
First, let's simplify the right side:
\(\sqrt{x^2-3x}=4x^2-12x-6\)
Notice that \(4x^2-12x\) can be factored as \(4(x^2-3x)\).
So, the equation becomes \(\sqrt{x^2-3x}=4(x^2-3x)-6\).
Let's substitute \(x^2-3x = t\). The equation then becomes:
\(\sqrt{t} = 4t - 6\)
To remove the square root, we square both sides of the equation:
\((\sqrt{t})^2 = (4t - 6)^2\)
\(t = (4t)^2 - 2(4t)(6) + 6^2\)
\(t = 16t^2 - 48t + 36\)
Rearrange into a standard quadratic equation by moving all terms to one side:
\(0 = 16t^2 - 48t - t + 36\)
\(16t^2 - 49t + 36 = 0\)
To solve this quadratic equation, we use the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
Here, \(a = 16\), \(b = -49\), \(c = 36\).
\(t = \frac{-(-49) \pm \sqrt{(-49)^2 - 4(16)(36)}}{2(16)}\)
\(t = \frac{49 \pm \sqrt{2401 - 2304}}{32}\)
\(t = \frac{49 \pm \sqrt{97}}{32}\)
So, we have two possible values for \(t\):
\(t = \frac{49 + \sqrt{97}}{32}\) or \(t = \frac{49 - \sqrt{97}}{32}\)
Now, we need to substitute back \(x^2-3x = t\).
Case-I: When \(t = \frac{49 + \sqrt{97}}{32}\)
\(x^2 - 3x = \frac{49 + \sqrt{97}}{32}\)
\(32x^2 - 96x = 49 + \sqrt{97}\)
\(32x^2 - 96x - (49 + \sqrt{97}) = 0\)
Using the quadratic formula for \(x\): \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
Here, \(a = 32\), \(b = -96\), \(c = -(49 + \sqrt{97})\).
\(x = \frac{-(-96) \pm \sqrt{(-96)^2 - 4(32)(-(49 + \sqrt{97}))}}{2(32)}\)
\(x = \frac{96 \pm \sqrt{9216 + 128(49 + \sqrt{97})}}{64}\)
\(x = \frac{96 \pm \sqrt{9216 + 6272 + 128\sqrt{97}}}{64}\)
\(x = \frac{96 \pm \sqrt{15488 + 128\sqrt{97}}}{64}\)

Case-II: When \(t = \frac{49 - \sqrt{97}}{32}\)
\(x^2 - 3x = \frac{49 - \sqrt{97}}{32}\)
\(32x^2 - 96x = 49 - \sqrt{97}\)
\(32x^2 - 96x - (49 - \sqrt{97}) = 0\)
Using the quadratic formula for \(x\): \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
Here, \(a = 32\), \(b = -96\), \(c = -(49 - \sqrt{97})\).
\(x = \frac{96 \pm \sqrt{9216 + 128(49 - \sqrt{97})}}{64}\)
\(x = \frac{96 \pm \sqrt{9216 + 6272 - 128\sqrt{97}}}{64}\)
\(x = \frac{96 \pm \sqrt{15488 - 128\sqrt{97}}}{64}\)
These solutions are quite complex. Let's re-examine the OCR's provided solution. The OCR solution for question 7 seems to be following a different interpretation of the problem. It starts with: `t = (4t – 3)² ⇒ t = 16t² – 24t + 9 ⇒ 16t² – 25t + 9 = 0` This means the equation it solved was \(\sqrt{t} = 4t - 3\). If the original question was \(\sqrt{x^2-3x}=4x^2-12x-3\), then substituting \(x^2-3x = t\) would lead to \(\sqrt{t} = 4t - 3\). Let's assume the question should have been: \(\sqrt{x^2-3x}=4x^2-12x-3\). Let the given equation be \(\sqrt{x^2-3x}=4x^2-12x-3\).
We can rewrite \(4x^2-12x\) as \(4(x^2-3x)\).
So, \(\sqrt{x^2-3x}=4(x^2-3x)-3\).
Let \(x^2-3x = t\). Then the equation becomes:
\(\sqrt{t} = 4t - 3\)
Square both sides to remove the square root:
\((\sqrt{t})^2 = (4t - 3)^2\)
\(t = 16t^2 - 24t + 9\)
Rearrange into a standard quadratic equation:
\(16t^2 - 24t - t + 9 = 0\)
\(16t^2 - 25t + 9 = 0\)
Now, use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) for \(a=16, b=-25, c=9\):
\(t = \frac{-(-25) \pm \sqrt{(-25)^2 - 4(16)(9)}}{2(16)}\)
\(t = \frac{25 \pm \sqrt{625 - 576}}{32}\)
\(t = \frac{25 \pm \sqrt{49}}{32}\)
\(t = \frac{25 \pm 7}{32}\)
So, we have two possible values for \(t\):
\(t = \frac{25 + 7}{32} = \frac{32}{32} = 1\)
or \(t = \frac{25 - 7}{32} = \frac{18}{32} = \frac{9}{16}\).
Now we substitute \(x^2-3x = t\) back.

Case-I: When \(t = 1\)
\(x^2 - 3x = 1\)
\(x^2 - 3x - 1 = 0\)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) for \(a=1, b=-3, c=-1\):
\(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)}\)
\(x = \frac{3 \pm \sqrt{9 + 4}}{2}\)
\(x = \frac{3 \pm \sqrt{13}}{2}\)

Case-II: When \(t = \frac{9}{16}\)
\(x^2 - 3x = \frac{9}{16}\)
Multiply by 16 to remove the fraction:
\(16x^2 - 48x = 9\)
\(16x^2 - 48x - 9 = 0\)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) for \(a=16, b=-48, c=-9\):
\(x = \frac{-(-48) \pm \sqrt{(-48)^2 - 4(16)(-9)}}{2(16)}\)
\(x = \frac{48 \pm \sqrt{2304 + 576}}{32}\)
\(x = \frac{48 \pm \sqrt{2880}}{32}\)
To simplify \(\sqrt{2880}\), we find its factors: \(2880 = 576 \times 5 = (24^2) \times 5\). So \(\sqrt{2880} = 24\sqrt{5}\). *Self-correction*: The OCR states `48±√64 x 45` then `48±8×3√5`. This means \(2880 = 64 \times 45 = (8^2) \times (9 \times 5) = 8^2 \times 3^2 \times 5\). So \(\sqrt{2880} = \sqrt{8^2 \times 3^2 \times 5} = 8 \times 3 \times \sqrt{5} = 24\sqrt{5}\). \(x = \frac{48 \pm 24\sqrt{5}}{32}\)
Divide the numerator and denominator by 8:
\(x = \frac{6 \pm 3\sqrt{5}}{4}\)
We must also check for extraneous roots when squaring both sides. For \(\sqrt{t} = 4t - 3\), we need \(t \ge 0\) and \(4t - 3 \ge 0 \implies t \ge \frac{3}{4}\). Both \(t=1\) and \(t=\frac{9}{16}\) satisfy \(t \ge \frac{3}{4}\) (since \(\frac{9}{16} = 0.5625\) and \(\frac{3}{4} = 0.75\), wait, \(t = \frac{9}{16}\) does not satisfy \(t \ge \frac{3}{4}\)). Since \(t = \frac{9}{16} < \frac{3}{4}\), the solutions from Case-II are extraneous roots because they do not satisfy the condition \(4t-3 \ge 0\). Therefore, only the solutions from Case-I are valid.
Thus, the solutions are \(x = \frac{3 \pm \sqrt{13}}{2}\). It's crucial to verify solutions when squaring equations.
In simple words: We first make the equation simpler by replacing \(x^2-3x\) with \(t\). Then we solve for \(t\), making sure to square both sides carefully. After finding \(t\), we solve for \(x\). We also check if our answers are real and make sense for the original equation.

🎯 Exam Tip: When you square both sides of an equation to remove a square root, it's essential to check your final solutions in the *original* equation to eliminate any extraneous (false) roots that might have been introduced by squaring.

Question 8. \(\sqrt{\frac{x^2+2}{x^2-2}}+6 \sqrt{\frac{x^2-2}{x^2+2}} = 5.\)
Answer: The given equation is \(\sqrt{\frac{x^2+2}{x^2-2}}+6 \sqrt{\frac{x^2-2}{x^2+2}} = 5\).
Notice that the two square root terms are reciprocals of each other. Let's make a substitution.
Let \(y = \sqrt{\frac{x^2+2}{x^2-2}}\). Then the second term is \(6 \times \frac{1}{y}\).
The equation becomes \(y + \frac{6}{y} = 5\).
Multiply the entire equation by \(y\) to clear the fraction (assuming \(y \neq 0\)):
\(y^2 + 6 = 5y\)
Rearrange into a standard quadratic equation:
\(y^2 - 5y + 6 = 0\)
Now, factorize this quadratic equation:
\(y^2 - 2y - 3y + 6 = 0\)
\(y(y - 2) - 3(y - 2) = 0\)
\((y - 2)(y - 3) = 0\)
This means either \(y - 2 = 0\) or \(y - 3 = 0\).
So, \(y = 2\) or \(y = 3\).
Now, we substitute back \(y = \sqrt{\frac{x^2+2}{x^2-2}}\).

Case-I: When \(y = 2\)
\(\sqrt{\frac{x^2+2}{x^2-2}} = 2\)
Square both sides to remove the square root:
\((\sqrt{\frac{x^2+2}{x^2-2}})^2 = 2^2\)
\(\frac{x^2+2}{x^2-2} = 4\)
Multiply both sides by \((x^2-2)\):
\(x^2 + 2 = 4(x^2 - 2)\)
\(x^2 + 2 = 4x^2 - 8\)
Move \(x^2\) terms to one side and constants to the other:
\(2 + 8 = 4x^2 - x^2\)
\(10 = 3x^2\)
\(x^2 = \frac{10}{3}\)
Take the square root of both sides:
\(x = \pm \sqrt{\frac{10}{3}}\)

Case-II: When \(y = 3\)
\(\sqrt{\frac{x^2+2}{x^2-2}} = 3\)
Square both sides:
\((\sqrt{\frac{x^2+2}{x^2-2}})^2 = 3^2\)
\(\frac{x^2+2}{x^2-2} = 9\)
Multiply both sides by \((x^2-2)\):
\(x^2 + 2 = 9(x^2 - 2)\)
\(x^2 + 2 = 9x^2 - 18\)
Move terms:
\(2 + 18 = 9x^2 - x^2\)
\(20 = 8x^2\)
\(x^2 = \frac{20}{8}\)
Simplify the fraction:
\(x^2 = \frac{5}{2}\)
Take the square root of both sides:
\(x = \pm \sqrt{\frac{5}{2}}\)
Thus, the solutions are \(x = \pm \sqrt{\frac{10}{3}}, \pm \sqrt{\frac{5}{2}}\). Recognizing reciprocal terms simplifies the problem significantly.
In simple words: We see that parts of the equation are upside down versions of each other. So we replace one of them with \(y\), which makes the equation much simpler. We solve for \(y\), then put the original terms back to find \(x\).

🎯 Exam Tip: Always look for reciprocal expressions within an equation. Substituting one of them as a new variable (and the other as its reciprocal) often reduces complex fractional equations to simpler quadratic forms.

Question 9. \(\sqrt{\frac{2x^2+1}{x^2-1}}+6 \sqrt{\frac{x^2-1}{2x^2+1}} = 5.\)
Answer: The given equation is \(\sqrt{\frac{2x^2+1}{x^2-1}}+6 \sqrt{\frac{x^2-1}{2x^2+1}} = 5\).
Similar to the previous problem, the two square root terms are reciprocals of each other.
Let \(t = \sqrt{\frac{2x^2+1}{x^2-1}}\). Then the second term is \(6 \times \frac{1}{t}\).
The equation becomes \(t + \frac{6}{t} = 5\).
Multiply the entire equation by \(t\) to clear the fraction (assuming \(t \neq 0\)):
\(t^2 + 6 = 5t\)
Rearrange into a standard quadratic equation:
\(t^2 - 5t + 6 = 0\)
Now, factorize this quadratic equation:
\(t^2 - 2t - 3t + 6 = 0\)
\(t(t - 2) - 3(t - 2) = 0\)
\((t - 2)(t - 3) = 0\)
This means either \(t - 2 = 0\) or \(t - 3 = 0\).
So, \(t = 2\) or \(t = 3\).
Now, we substitute back \(t = \sqrt{\frac{2x^2+1}{x^2-1}}\).

Case-I: When \(t = 2\)
\(\sqrt{\frac{2x^2+1}{x^2-1}} = 2\)
Square both sides to remove the square root:
\((\sqrt{\frac{2x^2+1}{x^2-1}})^2 = 2^2\)
\(\frac{2x^2+1}{x^2-1} = 4\)
Multiply both sides by \((x^2-1)\):
\(2x^2 + 1 = 4(x^2 - 1)\)
\(2x^2 + 1 = 4x^2 - 4\)
Move \(x^2\) terms to one side and constants to the other:
\(1 + 4 = 4x^2 - 2x^2\)
\(5 = 2x^2\)
\(x^2 = \frac{5}{2}\)
Take the square root of both sides:
\(x = \pm \sqrt{\frac{5}{2}}\)

Case-II: When \(t = 3\)
\(\sqrt{\frac{2x^2+1}{x^2-1}} = 3\)
Square both sides:
\((\sqrt{\frac{2x^2+1}{x^2-1}})^2 = 3^2\)
\(\frac{2x^2+1}{x^2-1} = 9\)
Multiply both sides by \((x^2-1)\):
\(2x^2 + 1 = 9(x^2 - 1)\)
\(2x^2 + 1 = 9x^2 - 9\)
Move terms:
\(1 + 9 = 9x^2 - 2x^2\)
\(10 = 7x^2\)
\(x^2 = \frac{10}{7}\)
Take the square root of both sides:
\(x = \pm \sqrt{\frac{10}{7}}\)
Thus, the solutions are \(x = \pm \sqrt{\frac{5}{2}}, \pm \sqrt{\frac{10}{7}}\). This pattern of substitution is very common in advanced algebra.
In simple words: We notice the two square root parts are opposites of each other. So we call one part \(t\), which makes the equation much simpler. We solve for \(t\), then put the original parts back to find the values for \(x\).

🎯 Exam Tip: Always look for terms that are reciprocals of each other, especially under square roots. Substituting one term as a variable (e.g., \(t\)) and the other as its reciprocal (\(1/t\)) often simplifies the problem into a standard quadratic equation.

Question 10. \(x(x - 1)(x + 2)(x - 3) + 8 = 0.\)
Answer: The given equation is \(x(x - 1)(x + 2)(x - 3) + 8 = 0\).
To solve this, we need to group the factors smartly so that they produce a common term when multiplied.
Let's group \((x)\) with \((x - 1)\) and \((x + 2)\) with \((x - 3)\).
\((x)(x - 1) = x^2 - x\)
\((x + 2)(x - 3) = x^2 - 3x + 2x - 6 = x^2 - x - 6\)
So, the equation becomes \((x^2 - x)(x^2 - x - 6) + 8 = 0\).
Now, we can make a substitution. Let \(t = x^2 - x\).
The equation becomes \(t(t - 6) + 8 = 0\).
Expand and rearrange into a standard quadratic equation:
\(t^2 - 6t + 8 = 0\)
Now, we factorize this quadratic equation:
\(t^2 - 4t - 2t + 8 = 0\)
\(t(t - 4) - 2(t - 4) = 0\)
\((t - 4)(t - 2) = 0\)
This means either \(t - 4 = 0\) or \(t - 2 = 0\).
So, \(t = 4\) or \(t = 2\).
Now, we substitute back \(x^2 - x = t\).

Case-I: When \(t = 4\)
\(x^2 - x = 4\)
\(x^2 - x - 4 = 0\)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) for \(a=1, b=-1, c=-4\):
\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)}\)
\(x = \frac{1 \pm \sqrt{1 + 16}}{2}\)
\(x = \frac{1 \pm \sqrt{17}}{2}\)

Case-II: When \(t = 2\)
\(x^2 - x = 2\)
\(x^2 - x - 2 = 0\)
Factorize this quadratic equation:
\(x^2 - 2x + x - 2 = 0\)
\(x(x - 2) + 1(x - 2) = 0\)
\((x + 1)(x - 2) = 0\)
This means either \(x + 1 = 0\) or \(x - 2 = 0\).
So, \(x = -1\) or \(x = 2\).
Thus, the solutions are \(x = -1, 2, \frac{1 \pm \sqrt{17}}{2}\). Careful grouping of terms is key to solving higher-degree polynomial equations.
In simple words: We group the terms in pairs that give us a common part. Then we replace this common part with \(t\) to make a simple quadratic equation. We solve for \(t\), then put the original expression back to find all the values for \(x\).

🎯 Exam Tip: For polynomial equations with four factors, group them strategically (e.g., first with fourth, second with third, if constants sum to same value) to create a common quadratic expression, which can then be substituted for simplification.

Question 11. \((x - 7)(x - 3)(x + 1)(x + 5) = 1680.\)
Answer: The given equation is \((x - 7)(x - 3)(x + 1)(x + 5) = 1680\).
To solve this, we need to group the factors smartly so that they produce a common term when multiplied.
Let's group \((x - 7)\) with \((x + 5)\) and \((x - 3)\) with \((x + 1)\) because the sums of their constants are equal (e.g., \(-7+5 = -2\) and \(-3+1 = -2\)).
\((x - 7)(x + 5) = x^2 + 5x - 7x - 35 = x^2 - 2x - 35\)
\((x - 3)(x + 1) = x^2 + x - 3x - 3 = x^2 - 2x - 3\)
So, the equation becomes \((x^2 - 2x - 35)(x^2 - 2x - 3) = 1680\).
Now, we can make a substitution. Let \(t = x^2 - 2x\).
The equation becomes \((t - 35)(t - 3) = 1680\).
Expand and rearrange into a standard quadratic equation:
\(t^2 - 3t - 35t + 105 = 1680\)
\(t^2 - 38t + 105 - 1680 = 0\)
\(t^2 - 38t - 1575 = 0\)
Now, use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) for \(a=1, b=-38, c=-1575\):
\(t = \frac{-(-38) \pm \sqrt{(-38)^2 - 4(1)(-1575)}}{2(1)}\)
\(t = \frac{38 \pm \sqrt{1444 + 6300}}{2}\)
\(t = \frac{38 \pm \sqrt{7744}}{2}\)
We calculate \(\sqrt{7744}\). It is 88.
\(t = \frac{38 \pm 88}{2}\)
So, we have two possible values for \(t\):
\(t = \frac{38 + 88}{2} = \frac{126}{2} = 63\)
or \(t = \frac{38 - 88}{2} = \frac{-50}{2} = -25\).
Now, we substitute back \(x^2 - 2x = t\).

Case-I: When \(t = 63\)
\(x^2 - 2x = 63\)
\(x^2 - 2x - 63 = 0\)
Factorize this quadratic equation:
\(x^2 - 9x + 7x - 63 = 0\)
\(x(x - 9) + 7(x - 9) = 0\)
\((x + 7)(x - 9) = 0\)
This means either \(x + 7 = 0\) or \(x - 9 = 0\).
So, \(x = -7\) or \(x = 9\).

Case-II: When \(t = -25\)
\(x^2 - 2x = -25\)
\(x^2 - 2x + 25 = 0\)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) for \(a=1, b=-2, c=25\):
\(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(25)}}{2(1)}\)
\(x = \frac{2 \pm \sqrt{4 - 100}}{2}\)
\(x = \frac{2 \pm \sqrt{-96}}{2}\)
Since the discriminant is negative (\(-96 < 0\)), there are no real solutions for this case. We can express them as complex numbers:
\(x = \frac{2 \pm \sqrt{96}i}{2} = \frac{2 \pm 4\sqrt{6}i}{2} = 1 \pm 2\sqrt{6}i\)
Thus, the real solutions are \(x = 9, -7\). Complex solutions are also possible depending on the discriminant.
In simple words: We smartly multiply pairs of terms to get a common expression. Then we replace this common expression with \(t\) to get a simple quadratic equation. We solve for \(t\), then put the original expression back to find all the \(x\) values, including any complex ones.

🎯 Exam Tip: For equations involving products of four linear factors, group them such that the sum of the constant terms in each pair is equal. This trick creates a common quadratic term for substitution.

Question 12. \((2x - 7)(x^2 - 9)(2x + 5) = 91.\)
Answer: The given equation is \((2x - 7)(x^2 - 9)(2x + 5) = 91\).
First, let's expand \(x^2 - 9\) using the difference of squares formula, \(a^2 - b^2 = (a-b)(a+b)\):
\(x^2 - 9 = (x - 3)(x + 3)\)
So, the equation becomes \((2x - 7)(x - 3)(x + 3)(2x + 5) = 91\).
Now, we group the factors to create common terms when multiplied. Let's group \((2x - 7)\) with \((x + 3)\) and \((x - 3)\) with \((2x + 5)\). This grouping is usually done such that the products form similar terms.
Let's try a different grouping for common factor: \((2x-7)\) with \((2x+5)\) and \((x-3)\) with \((x+3)\). The solution provided groups it as \((2x - 7)(x + 3)\) and \((x - 3)(2x + 5)\).
Let's follow that grouping:
\((2x - 7)(x + 3) = 2x^2 + 6x - 7x - 21 = 2x^2 - x - 21\)
\((x - 3)(2x + 5) = 2x^2 + 5x - 6x - 15 = 2x^2 - x - 15\)
So, the equation becomes \((2x^2 - x - 21)(2x^2 - x - 15) = 91\).
Now, we can make a substitution. Let \(t = 2x^2 - x\).
The equation becomes \((t - 21)(t - 15) = 91\).
Expand and rearrange into a standard quadratic equation:
\(t^2 - 15t - 21t + 315 = 91\)
\(t^2 - 36t + 315 - 91 = 0\)
\(t^2 - 36t + 224 = 0\)
Now, we factorize this quadratic equation:
\(t^2 - 8t - 28t + 224 = 0\)
\(t(t - 8) - 28(t - 8) = 0\)
\((t - 8)(t - 28) = 0\)
This means either \(t - 8 = 0\) or \(t - 28 = 0\).
So, \(t = 8\) or \(t = 28\).
Now, we substitute back \(2x^2 - x = t\).

Case-I: When \(t = 8\)
\(2x^2 - x = 8\)
\(2x^2 - x - 8 = 0\)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) for \(a=2, b=-1, c=-8\):
\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-8)}}{2(2)}\)
\(x = \frac{1 \pm \sqrt{1 + 64}}{4}\)
\(x = \frac{1 \pm \sqrt{65}}{4}\)

Case-II: When \(t = 28\)
\(2x^2 - x = 28\)
\(2x^2 - x - 28 = 0\)
Factorize this quadratic equation:
\(2x^2 - 8x + 7x - 28 = 0\)
\(2x(x - 4) + 7(x - 4) = 0\)
\((x - 4)(2x + 7) = 0\)
This means either \(x - 4 = 0\) or \(2x + 7 = 0\).
So, \(x = 4\) or \(x = -\frac{7}{2}\).
Thus, the solutions are \(x = 4, -\frac{7}{2}, \frac{1 \pm \sqrt{65}}{4}\). Factoring expressions correctly is crucial here.
In simple words: We first break down the given equation into simpler factors. Then we group these factors carefully so that when we multiply them, we get a common part. We replace this common part with \(t\) to make a simple quadratic equation, solve for \(t\), and then find all the values for \(x\).

🎯 Exam Tip: When given an equation with multiple factors, expand the factors in pairs that create a common quadratic expression (e.g., \(Ax^2+Bx\)) to enable effective substitution and simplification.

Question 13. By substituting \(y = 2^x\), or otherwise, solve the equation \(2^{2x} + 2^{x+2} - 4 \times 2^3 = 0.\)
Answer: The given equation is \(2^{2x} + 2^{x+2} - 4 \times 2^3 = 0\).
First, let's simplify the terms using exponent rules:
\(2^{2x} = (2^x)^2\)
\(2^{x+2} = 2^x \times 2^2 = 2^x \times 4\)
\(4 \times 2^3 = 4 \times 8 = 32\)
So the equation becomes \((2^x)^2 + 4 \times 2^x - 32 = 0\).
We are asked to substitute \(y = 2^x\).
The equation becomes \(y^2 + 4y - 32 = 0\).
Now, we factorize this quadratic equation:
\(y^2 + 8y - 4y - 32 = 0\)
\(y(y + 8) - 4(y + 8) = 0\)
\((y + 8)(y - 4) = 0\)
This means either \(y + 8 = 0\) or \(y - 4 = 0\).
So, \(y = -8\) or \(y = 4\).
Now, we substitute back \(y = 2^x\).
\(2^x = -8\) or \(2^x = 4\).
For \(2^x = -8\): Since \(2^x\) is always positive for any real value of \(x\), there is no real solution for \(2^x = -8\).
For \(2^x = 4\): We know that \(4 = 2^2\).
So, \(2^x = 2^2\)
Comparing the powers, we get \(x = 2\).
Thus, the only real solution is \(x = 2\). Remember that exponential functions with a positive base never yield negative values.
In simple words: We first make the equation simpler using rules of powers. Then, we replace \(2^x\) with \(y\), which turns the equation into a simple quadratic. We solve for \(y\), but only keep the positive answer for \(y\) because \(2^x\) cannot be negative. Finally, we find the value of \(x\).

🎯 Exam Tip: When solving exponential equations by substitution, always check the validity of your substituted variable's values. For instance, if \(t = a^x\) and \(a > 0\), then \(t\) must always be positive; discard any negative values for \(t\).

Question 14. \(2^{x^2} : 2^x = 8 : 1.\)
Answer: The given ratio equation is \(2^{x^2} : 2^x = 8 : 1\).
This can be written as a fraction:
\(\frac{2^{x^2}}{2^x} = \frac{8}{1}\)
Using the exponent rule \(\frac{a^m}{a^n} = a^{m-n}\) for the left side:
\(2^{x^2 - x} = 8\)
We know that \(8\) can be written as a power of 2: \(8 = 2^3\).
So, \(2^{x^2 - x} = 2^3\)
Since the bases are the same, we can equate the exponents:
\(x^2 - x = 3\)
Rearrange into a standard quadratic equation:
\(x^2 - x - 3 = 0\)
Now, use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) for \(a=1, b=-1, c=-3\):
\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)}\)
\(x = \frac{1 \pm \sqrt{1 + 12}}{2}\)
\(x = \frac{1 \pm \sqrt{13}}{2}\)
Thus, the solutions are \(x = \frac{1 \pm \sqrt{13}}{2}\). Expressing numbers with the same base is a fundamental step in solving exponential equations.
In simple words: We rewrite the ratio as a fraction and then use rules of powers to simplify it. We change 8 to \(2^3\). Then, because the bases are the same, we can make the powers equal and solve the simple quadratic equation for \(x\).

🎯 Exam Tip: When dealing with exponential equations involving ratios, convert them to fractional form. Then, use exponent rules to simplify and aim to express both sides with the same base to equate the powers.

Question 15. \(2^{2x+3} + 2^{x+3} = 1 + 2^x.\)
Answer: The given equation is \(2^{2x+3} + 2^{x+3} = 1 + 2^x\).
Let's simplify the terms using exponent rules, \(a^{m+n} = a^m \times a^n\):
\(2^{2x+3} = 2^{2x} \times 2^3 = (2^x)^2 \times 8\)
\(2^{x+3} = 2^x \times 2^3 = 2^x \times 8\)
The equation becomes \(8 \times (2^x)^2 + 8 \times 2^x = 1 + 2^x\).
Let's substitute \(t = 2^x\).
The equation becomes \(8t^2 + 8t = 1 + t\).
Rearrange into a standard quadratic equation:
\(8t^2 + 8t - t - 1 = 0\)
\(8t^2 + 7t - 1 = 0\)
Now, we factorize this quadratic equation:
\(8t^2 + 8t - t - 1 = 0\)
\(8t(t + 1) - 1(t + 1) = 0\)
\((t + 1)(8t - 1) = 0\)
This means either \(t + 1 = 0\) or \(8t - 1 = 0\).
So, \(t = -1\) or \(8t = 1 \implies t = \frac{1}{8}\).
Now, we substitute back \(t = 2^x\).
\(2^x = -1\) or \(2^x = \frac{1}{8}\).
For \(2^x = -1\): Since \(2^x\) is always positive for any real value of \(x\), there is no real solution for \(2^x = -1\).
For \(2^x = \frac{1}{8}\): We know that \(\frac{1}{8} = \frac{1}{2^3} = 2^{-3}\).
So, \(2^x = 2^{-3}\)
Comparing the powers, we get \(x = -3\).
Thus, the only real solution is \(x = -3\). Always remember to check for real solutions in exponential functions.
In simple words: First, we use rules of powers to simplify the terms. Then we replace \(2^x\) with \(t\), which changes the equation into a simple quadratic. We solve for \(t\), but we only keep the positive answer because \(2^x\) can't be negative. Finally, we find the value of \(x\).

🎯 Exam Tip: Carefully apply exponent rules to simplify terms before substitution. Be particularly mindful that \(a^x\) where \(a>0\) will always yield a positive result, so reject any negative values for the substituted variable.

Question 16. \(4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1}.\)
Answer: The given equation is \(4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1}\).
First, let's rewrite all terms using base exponents. We know \(4^x = (2^2)^x = 2^{2x}\).
Also, \(2^{2x-1} = 2^{2x} \times 2^{-1} = \frac{2^{2x}}{2}\).
And \(3^{x-\frac{1}{2}} = 3^x \times 3^{-\frac{1}{2}} = \frac{3^x}{\sqrt{3}}\).
And \(3^{x+\frac{1}{2}} = 3^x \times 3^{\frac{1}{2}} = 3^x \sqrt{3}\).
Substitute these simplified terms back into the original equation:
\(2^{2x} - \frac{3^x}{\sqrt{3}} = 3^x \sqrt{3} - \frac{2^{2x}}{2}\)
Now, gather terms with \(2^{2x}\) on one side and terms with \(3^x\) on the other side:
\(2^{2x} + \frac{2^{2x}}{2} = 3^x \sqrt{3} + \frac{3^x}{\sqrt{3}}\)
Factor out common terms:
\(2^{2x}(1 + \frac{1}{2}) = 3^x(\sqrt{3} + \frac{1}{\sqrt{3}})\)
Simplify the expressions in the parentheses:
\(1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2}\)
\(\sqrt{3} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{3+1}{\sqrt{3}} = \frac{4}{\sqrt{3}}\)
So, the equation becomes:
\(2^{2x} \times \frac{3}{2} = 3^x \times \frac{4}{\sqrt{3}}\)
Rearrange the terms to group \(2^{2x}\) and \(3^x\):
\(\frac{2^{2x}}{3^x} = \frac{4}{\sqrt{3}} \times \frac{2}{3}\)
\(\frac{(2^2)^x}{3^x} = \frac{8}{3\sqrt{3}}\)
\(\frac{4^x}{3^x} = \frac{8}{3\sqrt{3}}\)
\(\left(\frac{4}{3}\right)^x = \frac{8}{3\sqrt{3}}\)
We can rewrite the right side with powers of 2 and 3:
\(\frac{8}{3\sqrt{3}} = \frac{2^3}{3^1 \times 3^{1/2}} = \frac{2^3}{3^{3/2}}\)
\(\frac{2^3}{3^{3/2}} = \frac{2^3}{(3^{1/2})^3} = \frac{2^3}{(\sqrt{3})^3} = \left(\frac{2}{\sqrt{3}}\right)^3\)
However, the OCR solution proceeds differently at this step. It seems to interpret \(\frac{8}{3\sqrt{3}}\) as \(\frac{4^{3/2}}{3^{3/2}}\).
Let's follow the OCR's implied steps:
\(\frac{8}{3\sqrt{3}} = \frac{2^3}{3 \times \sqrt{3}} = \frac{2^3}{3^{3/2}}\)
This can be written as \(\left(\frac{2}{3^{1/2}}\right)^3 = \left(\frac{2}{\sqrt{3}}\right)^3\).
So, \(\left(\frac{4}{3}\right)^x = \left(\frac{2}{\sqrt{3}}\right)^3\)
This form does not allow direct comparison of bases. Let's re-check the OCR solution part `43/2` and `33/2`. The OCR shows: `8/3x√3 = 4^3/2 / 3^3/2`. This would mean `8 = 4^3/2 = (2^2)^3/2 = 2^3`. This is correct. And `3√3 = 3^1 * 3^1/2 = 3^3/2`. This is also correct.
So, \(\left(\frac{4}{3}\right)^x = \left(\frac{4}{3}\right)^{3/2}\)
Wait, \(\frac{4^{3/2}}{3^{3/2}} = \left(\frac{4}{3}\right)^{3/2}\).
So, we have \(\left(\frac{4}{3}\right)^x = \left(\frac{4}{3}\right)^{3/2}\).
Since the bases are the same, we can equate the exponents:
\(x = \frac{3}{2}\)
Thus, the solution is \(x = \frac{3}{2}\). It's always beneficial to express both sides of an equation with similar bases before comparing powers.
In simple words: We rewrite all parts of the equation using simple power rules. We collect terms with the same base on each side. After simplifying both sides, we make sure they look alike, like \((A/B)^x = (A/B)^N\), so we can just say \(x = N\).

🎯 Exam Tip: When an equation involves terms with different bases (e.g., 2, 3, 4), try to express all terms using the smallest prime bases possible. Then, group similar base terms to simplify and solve for x.