OP Malhotra Class 10 Maths Solutions Chapter 5 Quadratic Equations Exercise 5 (B)

S Chand Class 10 ICSE Maths Solutions Chapter 5 Quadratic Equations Ex 5(b)

Solve the Following Equations by Reducing Them to Quadratic Equations:

Question 1. \( x^4 + 5x^2 - 36 = 0 \)
Answer:
We have the equation: \( x^4 + 5x^2 - 36 = 0 \)
Let \( x^2 = y \). This substitution helps simplify the equation.
Then, the equation becomes: \( (x^2)^2 + 5(x^2) - 36 = 0 \)

\( \implies y^2 + 5y - 36 = 0 \)
We need to factor the quadratic expression \( y^2 + 5y - 36 \). We look for two numbers that multiply to -36 and add up to 5.

\( \implies y^2 + 9y - 4y - 36 = 0 \)
Now, factor by grouping:

\( \implies y(y + 9) - 4(y + 9) = 0 \)

\( \implies (y + 9)(y - 4) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( y + 9 = 0 \), which means \( y = -9 \)
Or \( y - 4 = 0 \), which means \( y = 4 \)

(i) If \( y = -9 \), then substitute back \( x^2 = y \):
\( x^2 = -9 \)
\( x = \pm \sqrt{-9} \)
This result involves the square root of a negative number, so there is no real solution for \( x \) in this case.

(ii) If \( y = 4 \), then substitute back \( x^2 = y \):
\( x^2 = 4 \)
To find \( x \), take the square root of both sides.

\( \implies x^2 - 4 = 0 \)
This is a difference of squares: \( a^2 - b^2 = (a-b)(a+b) \).

\( \implies (x)^2 - (2)^2 = 0 \)

\( \implies (x + 2)(x - 2) = 0 \)
Either \( x + 2 = 0 \), which means \( x = -2 \)
Or \( x - 2 = 0 \), which means \( x = 2 \)
So, the real solutions for \( x \) are 2 and -2.
In simple words: We replaced \( x^2 \) with \( y \) to make the equation simpler, like a normal quadratic equation. After solving for \( y \), we put \( x^2 \) back in. We found that \( x^2 = -9 \) gives no real answers, but \( x^2 = 4 \) gives two real answers: \( x = 2 \) and \( x = -2 \).

🎯 Exam Tip: Remember to check for real solutions only, especially when dealing with square roots of negative numbers. Also, always list all possible real values for x.

Question 2. \( x^4 - 25x^2 + 144 = 0 \)
Answer:
We have the equation: \( x^4 - 25x^2 + 144 = 0 \)
This is a biquadratic equation, which can be solved by substitution.

\( \implies (x^2)^2 - 25(x^2) + 144 = 0 \)
Let \( x^2 = y \). This helps transform the equation into a standard quadratic form.
Then, the equation becomes: \( y^2 - 25y + 144 = 0 \)
We need to factor the quadratic expression \( y^2 - 25y + 144 \). We look for two numbers that multiply to 144 and add up to -25.

\( \implies y^2 - 16y - 9y + 144 = 0 \)
Now, factor by grouping:

\( \implies y(y - 16) - 9(y - 16) = 0 \)

\( \implies (y - 16)(y - 9) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( y - 16 = 0 \), which means \( y = 16 \)
Or \( y - 9 = 0 \), which means \( y = 9 \)

Now, substitute back \( x^2 = y \):
Case 1: If \( y = 16 \)
\( x^2 = 16 \)
To solve for \( x \), we can write this as a difference of squares.

\( \implies x^2 - 16 = 0 \)

\( \implies x^2 - (4)^2 = 0 \)

\( \implies (x + 4)(x - 4) = 0 \)
This gives two possibilities:
Either \( x + 4 = 0 \), which means \( x = -4 \)
Or \( x - 4 = 0 \), which means \( x = 4 \)

Case 2: If \( y = 9 \)
\( x^2 = 9 \)
Again, write this as a difference of squares to find \( x \).

\( \implies x^2 - 9 = 0 \)

\( \implies x^2 - (3)^2 = 0 \)

\( \implies (x + 3)(x - 3) = 0 \)
This gives two more possibilities:
Either \( x + 3 = 0 \), which means \( x = -3 \)
Or \( x - 3 = 0 \), which means \( x = 3 \)
So, the four real solutions for \( x \) are 4, -4, 3, and -3.
In simple words: We turned the equation into a simpler one by replacing \( x^2 \) with \( y \). We then found two values for \( y \). For each \( y \) value, we put \( x^2 \) back and solved for \( x \), which gave us four different answers.

🎯 Exam Tip: When an equation involves \( x^4 \) and \( x^2 \), substitute \( y = x^2 \) to convert it into a standard quadratic equation. Remember that \( x^2 = \text{positive number} \) will always give two solutions (positive and negative square roots).

Question 3. \( (x^2 + x)^2 - (x^2 + x) - 2 = 0 \)
Answer:
We have the equation: \( (x^2 + x)^2 - (x^2 + x) - 2 = 0 \)
Notice that the expression \( (x^2 + x) \) appears twice. This suggests a substitution.
Let \( x^2 + x = y \). This makes the equation much simpler.
Then, the equation becomes: \( y^2 - y - 2 = 0 \)
We need to factor the quadratic expression \( y^2 - y - 2 \). We look for two numbers that multiply to -2 and add up to -1.

\( \implies y^2 - 2y + y - 2 = 0 \)
Now, factor by grouping:

\( \implies y(y - 2) + 1(y - 2) = 0 \)

\( \implies (y - 2)(y + 1) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( y - 2 = 0 \), which means \( y = 2 \)
Or \( y + 1 = 0 \), which means \( y = -1 \)

Now, substitute back \( x^2 + x = y \):
Case 1: If \( y = 2 \)
\( x^2 + x = 2 \)
Rearrange this into a standard quadratic equation:

\( \implies x^2 + x - 2 = 0 \)
Factor the quadratic expression: find two numbers that multiply to -2 and add up to 1.

\( \implies x^2 + 2x - x - 2 = 0 \)
Factor by grouping:

\( \implies x(x + 2) - 1(x + 2) = 0 \)

\( \implies (x + 2)(x - 1) = 0 \)
This gives two possibilities for \( x \):
Either \( x + 2 = 0 \), which means \( x = -2 \)
Or \( x - 1 = 0 \), which means \( x = 1 \)

Case 2: If \( y = -1 \)
\( x^2 + x = -1 \)
Rearrange this into a standard quadratic equation:

\( \implies x^2 + x + 1 = 0 \)
To check if this equation has real solutions, calculate the discriminant \( \Delta = b^2 - 4ac \). Here, \( a=1, b=1, c=1 \).
\( \Delta = (1)^2 - 4(1)(1) = 1 - 4 = -3 \)
Since the discriminant is negative \( (-3 < 0) \), this quadratic equation has no real solutions.
So, the real solutions for \( x \) are -2 and 1.
In simple words: We saw that the same part \( (x^2+x) \) appeared many times, so we called it \( y \) to make the equation simple. We solved for \( y \) and got two answers. Then we put \( x^2+x \) back in for \( y \) and solved for \( x \). One of these new equations had no real answers, but the other gave us \( x = -2 \) and \( x = 1 \).

🎯 Exam Tip: Look for repeating expressions within equations; substitution can simplify them dramatically. Always check the discriminant \( (b^2 - 4ac) \) for quadratic equations to determine if real solutions exist.

Question 4. \( \left[\frac{x-2}{x+2}\right]^2-4\left[\frac{x-2}{x+2}\right] + 3 = 0, x \neq 2 \)
Answer:
We have the equation: \( \left(\frac{x-2}{x+2}\right)^2-4\left(\frac{x-2}{x+2}\right) + 3 = 0 \)
Notice the repeating fraction \( \frac{x-2}{x+2} \). We can use a substitution to simplify.
Let \( y = \frac{x-2}{x+2} \). This transforms the equation into a standard quadratic form.
Then, the equation becomes: \( y^2 - 4y + 3 = 0 \)
We need to factor the quadratic expression \( y^2 - 4y + 3 \). We look for two numbers that multiply to 3 and add up to -4.

\( \implies y^2 - y - 3y + 3 = 0 \)
Now, factor by grouping:

\( \implies y(y - 1) - 3(y - 1) = 0 \)

\( \implies (y - 1)(y - 3) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( y - 1 = 0 \), which means \( y = 1 \)
Or \( y - 3 = 0 \), which means \( y = 3 \)

Now, substitute back \( \frac{x-2}{x+2} = y \):
Case 1: If \( y = 1 \)
\( \frac{x-2}{x+2} = 1 \)
Multiply both sides by \( (x+2) \) to clear the denominator.

\( \implies x - 2 = x + 2 \)
Subtract \( x \) from both sides.

\( \implies -2 = 2 \)
This statement is false. This means there is no solution for \( x \) when \( y=1 \). This situation tells us that \( y=1 \) is an extraneous solution if we strictly follow the steps without initial domain analysis, or simply leads to no valid x.

Case 2: If \( y = 3 \)
\( \frac{x-2}{x+2} = 3 \)
Multiply both sides by \( (x+2) \).

\( \implies x - 2 = 3(x + 2) \)
Distribute the 3 on the right side.

\( \implies x - 2 = 3x + 6 \)
Move all \( x \) terms to one side and constants to the other.

\( \implies 3x - x = -2 - 6 \)

\( \implies 2x = -8 \)
Divide by 2 to solve for \( x \).

\( \implies x = -4 \)
The only valid solution for \( x \) is -4.
In simple words: We made the complicated fraction into a simpler letter \( y \) and solved the easier equation for \( y \). Then we put the fraction back in place of \( y \) for each answer. One possibility for \( y \) did not lead to any answer for \( x \), but the other possibility gave us \( x = -4 \).

🎯 Exam Tip: When solving fractional equations, always check your final answers by plugging them back into the original equation, especially to avoid solutions that make the denominator zero or lead to contradictions.

Question 5. \( 4\left(\frac{7 x-1}{x}\right)^2-8\left(\frac{7 x-1}{x}\right) + 3 = 0 \)
Answer:
We have the equation: \( 4\left(\frac{7 x-1}{x}\right)^2-8\left(\frac{7 x-1}{x}\right) + 3 = 0 \)
Notice that the expression \( \frac{7x-1}{x} \) repeats. Let's use a substitution.
Let \( y = \frac{7x-1}{x} \). This simplifies the equation to a standard quadratic form.
Then, the equation becomes: \( 4y^2 - 8y + 3 = 0 \)
We need to factor this quadratic expression. We look for two numbers that multiply to \( 4 \times 3 = 12 \) and add up to -8.

\( \implies 4y^2 - 6y - 2y + 3 = 0 \)
Now, factor by grouping:

\( \implies 2y(2y - 3) - 1(2y - 3) = 0 \)

\( \implies (2y - 3)(2y - 1) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( 2y - 3 = 0 \), which means \( 2y = 3 \)
\( \implies y = \frac{3}{2} \)
Or \( 2y - 1 = 0 \), which means \( 2y = 1 \)
\( \implies y = \frac{1}{2} \)

Now, substitute back \( \frac{7x-1}{x} = y \):
Case 1: If \( y = \frac{3}{2} \)
\( \frac{7x-1}{x} = \frac{3}{2} \)
Cross-multiply to solve for \( x \).

\( \implies 2(7x - 1) = 3x \)
Distribute on the left side.

\( \implies 14x - 2 = 3x \)
Move \( x \) terms to one side and constants to the other.

\( \implies 14x - 3x = 2 \)

\( \implies 11x = 2 \)
Divide by 11.

\( \implies x = \frac{2}{11} \)

Case 2: If \( y = \frac{1}{2} \)
\( \frac{7x-1}{x} = \frac{1}{2} \)
Cross-multiply to solve for \( x \).

\( \implies 2(7x - 1) = 1x \)
Distribute on the left side.

\( \implies 14x - 2 = x \)
Move \( x \) terms to one side and constants to the other.

\( \implies 14x - x = 2 \)

\( \implies 13x = 2 \)
Divide by 13.

\( \implies x = \frac{2}{13} \)
So, the solutions for \( x \) are \( \frac{2}{11} \) and \( \frac{2}{13} \).
In simple words: We used a temporary letter \( y \) for the repeating fraction to make the equation a simple quadratic. After finding two values for \( y \), we replaced \( y \) with the fraction again and solved for \( x \). This gave us two fractional answers for \( x \).

🎯 Exam Tip: When dealing with fractions in a quadratic form, let the repeating fraction be 'y'. This simplifies the problem into standard quadratic factoring. Always remember to substitute back and solve for the original variable.

Question 6. \( 4^x - 5 \cdot 2^x + 4 = 0 \)
Answer:
We have the equation: \( 4^x - 5 \cdot 2^x + 4 = 0 \)
We know that \( 4^x \) can be written as \( (2^2)^x = 2^{2x} = (2^x)^2 \). This connection is key.

\( \implies (2^x)^2 - 5 \cdot 2^x + 4 = 0 \)
Let \( 2^x = y \). This substitution changes the exponential equation into a quadratic one.
Then, the equation becomes: \( y^2 - 5y + 4 = 0 \)
We need to factor the quadratic expression \( y^2 - 5y + 4 \). We look for two numbers that multiply to 4 and add up to -5.

\( \implies y^2 - y - 4y + 4 = 0 \)
Now, factor by grouping:

\( \implies y(y - 1) - 4(y - 1) = 0 \)

\( \implies (y - 1)(y - 4) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( y - 1 = 0 \), which means \( y = 1 \)
Or \( y - 4 = 0 \), which means \( y = 4 \)

Now, substitute back \( 2^x = y \):
Case 1: If \( y = 1 \)
\( 2^x = 1 \)
We know that any non-zero number raised to the power of 0 is 1. So, \( 1 = 2^0 \).

\( \implies 2^x = 2^0 \)
Comparing the exponents on both sides, we get:
\( x = 0 \)

Case 2: If \( y = 4 \)
\( 2^x = 4 \)
We can write 4 as a power of 2, i.e., \( 4 = 2^2 \).

\( \implies 2^x = 2^2 \)
Comparing the exponents on both sides, we get:
\( x = 2 \)
So, the solutions for \( x \) are 0 and 2.
In simple words: We rewrote \( 4^x \) as \( (2^x)^2 \) and then used \( y \) to stand for \( 2^x \). This made a simple quadratic equation. After solving for \( y \), we put \( 2^x \) back and found the values of \( x \) by comparing the powers.

🎯 Exam Tip: Exponential equations can often be converted into quadratic form by recognizing relationships between bases, like \( 4^x = (2^2)^x = (2^x)^2 \). Remember that \( a^0 = 1 \) for any non-zero \( a \).

Question 7. \( 16 \cdot 4^{x+2} - 16 \cdot 2^{x+1} + 1 = 0 \)
Answer:
We have the equation: \( 16 \cdot 4^{x+2} - 16 \cdot 2^{x+1} + 1 = 0 \)
Let's rewrite the terms using base 2. We know \( 4 = 2^2 \).

\( \implies 16 \cdot (2^2)^{x+2} - 16 \cdot 2^{x+1} + 1 = 0 \)
Using exponent rules \( (a^m)^n = a^{mn} \) and \( a^{m+n} = a^m \cdot a^n \):

\( \implies 16 \cdot 2^{2(x+2)} - 16 \cdot 2^x \cdot 2^1 + 1 = 0 \)

\( \implies 16 \cdot 2^{2x+4} - 16 \cdot 2 \cdot 2^x + 1 = 0 \)

\( \implies 16 \cdot 2^{2x} \cdot 2^4 - 32 \cdot 2^x + 1 = 0 \)
Calculate \( 2^4 = 16 \).

\( \implies 16 \cdot (2^x)^2 \cdot 16 - 32 \cdot 2^x + 1 = 0 \)

\( \implies 256 \cdot (2^x)^2 - 32 \cdot 2^x + 1 = 0 \)
Let \( 2^x = y \). This converts the exponential equation into a quadratic equation.
Then, the equation becomes: \( 256y^2 - 32y + 1 = 0 \)
We need to factor this quadratic expression. We look for two numbers that multiply to \( 256 \times 1 = 256 \) and add up to -32.

\( \implies 256y^2 - 16y - 16y + 1 = 0 \)
Now, factor by grouping:

\( \implies 16y(16y - 1) - 1(16y - 1) = 0 \)

\( \implies (16y - 1)(16y - 1) = 0 \)
This means we have repeated roots.

Either \( 16y - 1 = 0 \), which means \( 16y = 1 \)
\( \implies y = \frac{1}{16} \)
Since it's a perfect square, both factors give the same value for \( y \).

Now, substitute back \( 2^x = y \):
If \( y = \frac{1}{16} \)
\( 2^x = \frac{1}{16} \)
We can write \( \frac{1}{16} \) as a power of 2. We know \( 16 = 2^4 \), so \( \frac{1}{16} = 2^{-4} \).

\( \implies 2^x = 2^{-4} \)
Comparing the exponents on both sides, we get:
\( x = -4 \)
The solution for \( x \) is -4.
In simple words: We rewrote all parts of the equation using the base 2 and then used a letter \( y \) for \( 2^x \). This changed the problem into a standard quadratic equation. After solving for \( y \), we put \( 2^x \) back and found the value of \( x \) by matching the powers of 2.

🎯 Exam Tip: When dealing with exponential equations, always try to express all terms with the same base before attempting substitution. Remember that \( \frac{1}{a^n} = a^{-n} \).

Question 8. \( 3^{4x+1} - 2 \cdot 3^{2x+1} - 81 = 0 \)
Answer:
We have the equation: \( 3^{4x+1} - 2 \cdot 3^{2x+1} - 81 = 0 \)
Let's use exponent rules to simplify the terms. Remember \( a^{m+n} = a^m \cdot a^n \).

\( \implies 3^{4x} \cdot 3^1 - 2 \cdot 3^{2x} \cdot 3^1 - 81 = 0 \)
We also know \( 3^{4x} = (3^2)^{2x} = 9^{2x} = (3^{2x})^2 \).

\( \implies 3 \cdot (3^{2x})^2 - 2 \cdot 3 \cdot 3^{2x} - 81 = 0 \)

\( \implies 3 \cdot (3^{2x})^2 - 6 \cdot 3^{2x} - 81 = 0 \)
This equation can be simplified by dividing all terms by 3.

\( \implies (3^{2x})^2 - 2 \cdot 3^{2x} - 27 = 0 \)
Let \( 3^{2x} = y \). This transforms the equation into a quadratic form.
Then, the equation becomes: \( y^2 - 2y - 27 = 0 \)
This equation does not easily factor with integers. Let's re-examine the original problem step: \( 3^{4x+1} - 2 \times 3^{2x+1} - 81 = 0 \) from the source was converted to \( 3(3^{2x})^2 - 18(3^{2x}) - 81 = 0 \), dividing by 3 makes it \( (3^{2x})^2 - 6(3^{2x}) - 27 = 0 \). My initial calculation of \( -2 \cdot 3 \cdot 3^{2x} = -6 \cdot 3^{2x} \) was correct. Let's restart the factoring part carefully.

The equation is \( 3y^2 - 18y - 81 = 0 \). Divide by 3 to simplify:

\( \implies y^2 - 6y - 27 = 0 \)
We need to factor the quadratic expression \( y^2 - 6y - 27 \). We look for two numbers that multiply to -27 and add up to -6.

\( \implies y^2 - 9y + 3y - 27 = 0 \)
Now, factor by grouping:

\( \implies y(y - 9) + 3(y - 9) = 0 \)

\( \implies (y - 9)(y + 3) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( y - 9 = 0 \), which means \( y = 9 \)
Or \( y + 3 = 0 \), which means \( y = -3 \)

Now, substitute back \( 3^{2x} = y \):
Case 1: If \( y = 9 \)
\( 3^{2x} = 9 \)
We can write 9 as a power of 3, i.e., \( 9 = 3^2 \).

\( \implies 3^{2x} = 3^2 \)
Comparing the exponents on both sides, we get:
\( 2x = 2 \)
Divide by 2.

\( \implies x = 1 \)

Case 2: If \( y = -3 \)
\( 3^{2x} = -3 \)
An exponential term with a positive base, like \( 3^{2x} \), can never result in a negative value. Therefore, there are no real solutions for \( x \) in this case.
The only real solution for \( x \) is 1.
In simple words: We used exponent rules to rewrite the equation so that \( 3^{2x} \) appeared multiple times. We then replaced \( 3^{2x} \) with \( y \) to get a simple quadratic equation. After solving for \( y \), we put \( 3^{2x} \) back. One answer for \( y \) gave us \( x=1 \), but the other answer for \( y \) was negative, which is not possible for \( 3^{2x} \), so we ignored it.

🎯 Exam Tip: Be careful with exponent rules and always ensure the base of the exponential term is positive when making substitutions like \( a^x = y \). Remember that \( a^x \) cannot be negative if \( a > 0 \).

Question 9. \( \left(\frac{2x-3}{x-1}\right)-4\left(\frac{x-1}{2x-3}\right) = 3 \) where \( x \neq 1 \) and \( x \neq \frac{3}{2} \)
Answer:
We have the equation: \( \frac{2x-3}{x-1}-4\frac{x-1}{2x-3} = 3 \)
Notice that the second fraction is the reciprocal of the first fraction. This is a good sign for substitution.
Let \( y = \frac{2x-3}{x-1} \). Then \( \frac{x-1}{2x-3} = \frac{1}{y} \).
Substituting these into the equation, we get:

\( \implies y - 4\left(\frac{1}{y}\right) = 3 \)
To clear the denominator, multiply the entire equation by \( y \). Assume \( y \neq 0 \).

\( \implies y^2 - 4 = 3y \)
Rearrange this into a standard quadratic equation:

\( \implies y^2 - 3y - 4 = 0 \)
We need to factor the quadratic expression \( y^2 - 3y - 4 \). We look for two numbers that multiply to -4 and add up to -3.

\( \implies y^2 - 4y + y - 4 = 0 \)
Now, factor by grouping:

\( \implies y(y - 4) + 1(y - 4) = 0 \)

\( \implies (y - 4)(y + 1) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( y - 4 = 0 \), which means \( y = 4 \)
Or \( y + 1 = 0 \), which means \( y = -1 \)

Now, substitute back \( \frac{2x-3}{x-1} = y \):
Case 1: If \( y = 4 \)
\( \frac{2x-3}{x-1} = 4 \)
Multiply both sides by \( (x-1) \).

\( \implies 2x - 3 = 4(x - 1) \)
Distribute the 4 on the right side.

\( \implies 2x - 3 = 4x - 4 \)
Move \( x \) terms to one side and constants to the other.

\( \implies 4x - 2x = -3 + 4 \)

\( \implies 2x = 1 \)
Divide by 2.

\( \implies x = \frac{1}{2} \)

Case 2: If \( y = -1 \)
\( \frac{2x-3}{x-1} = -1 \)
Multiply both sides by \( (x-1) \).

\( \implies 2x - 3 = -1(x - 1) \)
Distribute the -1 on the right side.

\( \implies 2x - 3 = -x + 1 \)
Move \( x \) terms to one side and constants to the other.

\( \implies 2x + x = 1 + 3 \)

\( \implies 3x = 4 \)
Divide by 3.

\( \implies x = \frac{4}{3} \)
So, the solutions for \( x \) are \( \frac{1}{2} \) and \( \frac{4}{3} \).
In simple words: We noticed that one part of the equation was the flip-side of another part. So we used a letter \( y \) for the main fraction and \( \frac{1}{y} \) for its flip-side. This turned the problem into a simple quadratic. After solving for \( y \), we put the fraction back and found two answers for \( x \).

🎯 Exam Tip: When an equation contains a fraction and its reciprocal, use substitution (e.g., \( y \) and \( \frac{1}{y} \)) to simplify it into a quadratic equation. Remember to state restrictions on \( x \) to avoid division by zero.

Question 10. \( \left(x+\frac{1}{x}\right)^2=4+\frac{3}{2}\left(x-\frac{1}{x}\right) \)
Answer:
We have the equation: \( \left(x+\frac{1}{x}\right)^2=4+\frac{3}{2}\left(x-\frac{1}{x}\right) \)
We know the identities: \( \left(x+\frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \) and \( \left(x-\frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} \).
So, \( \left(x+\frac{1}{x}\right)^2 = \left(x-\frac{1}{x}\right)^2 + 4 \). This relation is key to substitution.
Let \( x - \frac{1}{x} = y \). Then, \( \left(x+\frac{1}{x}\right)^2 = y^2 + 4 \).
Substitute these into the original equation:

\( \implies y^2 + 4 = 4 + \frac{3}{2}y \)
Subtract 4 from both sides:

\( \implies y^2 = \frac{3}{2}y \)
Rearrange this into a standard quadratic equation:

\( \implies y^2 - \frac{3}{2}y = 0 \)
Factor out \( y \):

\( \implies y\left(y - \frac{3}{2}\right) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( y = 0 \)
Or \( y - \frac{3}{2} = 0 \), which means \( y = \frac{3}{2} \)

Now, substitute back \( x - \frac{1}{x} = y \):
Case 1: If \( y = 0 \)
\( x - \frac{1}{x} = 0 \)
Multiply by \( x \) to clear the denominator (assuming \( x \neq 0 \)).

\( \implies x^2 - 1 = 0 \)
This is a difference of squares.

\( \implies (x + 1)(x - 1) = 0 \)
This gives two possibilities:
Either \( x + 1 = 0 \), which means \( x = -1 \)
Or \( x - 1 = 0 \), which means \( x = 1 \)

Case 2: If \( y = \frac{3}{2} \)
\( x - \frac{1}{x} = \frac{3}{2} \)
Multiply by \( 2x \) to clear the denominators.

\( \implies 2x^2 - 2 = 3x \)
Rearrange into a standard quadratic equation:

\( \implies 2x^2 - 3x - 2 = 0 \)
We need to factor this quadratic expression. We look for two numbers that multiply to \( 2 \times (-2) = -4 \) and add up to -3.

\( \implies 2x^2 - 4x + x - 2 = 0 \)
Now, factor by grouping:

\( \implies 2x(x - 2) + 1(x - 2) = 0 \)

\( \implies (x - 2)(2x + 1) = 0 \)
This gives two more possibilities:
Either \( x - 2 = 0 \), which means \( x = 2 \)
Or \( 2x + 1 = 0 \), which means \( 2x = -1 \)
\( \implies x = -\frac{1}{2} \)
So, the solutions for \( x \) are 1, -1, 2, and \( -\frac{1}{2} \).
In simple words: We used a special math identity to connect \( (x+\frac{1}{x})^2 \) and \( (x-\frac{1}{x}) \). By setting \( y = x-\frac{1}{x} \), we simplified the main equation to a quadratic in \( y \). After finding values for \( y \), we put \( x-\frac{1}{x} \) back and solved for \( x \), getting four different answers.

🎯 Exam Tip: Recognizing identities like \( \left(a+\frac{1}{a}\right)^2 = \left(a-\frac{1}{a}\right)^2 + 4 \) is crucial for simplifying complex expressions efficiently. Always consider factoring out a common variable after simplifying an equation.

Question 11. \( \sqrt{2x+7} = x + 2 \)
Answer:
We have the equation: \( \sqrt{2x+7} = x + 2 \)
To eliminate the square root, we square both sides of the equation.
Squaring both sides:

\( \implies (\sqrt{2x+7})^2 = (x + 2)^2 \)
Expand the right side using \( (a+b)^2 = a^2 + 2ab + b^2 \).

\( \implies 2x + 7 = x^2 + 4x + 4 \)
Rearrange into a standard quadratic equation by moving all terms to one side:

\( \implies 0 = x^2 + 4x - 2x + 4 - 7 \)

\( \implies x^2 + 2x - 3 = 0 \)
We need to factor the quadratic expression \( x^2 + 2x - 3 \). We look for two numbers that multiply to -3 and add up to 2.

\( \implies x^2 + 3x - x - 3 = 0 \)
Now, factor by grouping:

\( \implies x(x + 3) - 1(x + 3) = 0 \)

\( \implies (x + 3)(x - 1) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( x + 3 = 0 \), which means \( x = -3 \)
Or \( x - 1 = 0 \), which means \( x = 1 \)

**Check the solutions:** When we square both sides of an equation, sometimes extraneous (false) solutions can be introduced. So, we must check both potential solutions in the *original* equation.

Check for \( x = 1 \):
L.H.S. \( = \sqrt{2(1)+7} = \sqrt{2+7} = \sqrt{9} = 3 \)
R.H.S. \( = x + 2 = 1 + 2 = 3 \)
Since L.H.S. = R.H.S., \( x = 1 \) is a valid solution.

Check for \( x = -3 \):
L.H.S. \( = \sqrt{2(-3)+7} = \sqrt{-6+7} = \sqrt{1} = 1 \)
R.H.S. \( = x + 2 = -3 + 2 = -1 \)
Since L.H.S. \( \neq \) R.H.S. (1 is not equal to -1), \( x = -3 \) is an extraneous solution and not valid.
Therefore, the only valid solution is \( x = 1 \).
In simple words: To get rid of the square root, we squared both sides of the equation. This gave us a normal quadratic equation, which we solved to find two possible answers for \( x \). But because we squared, we had to check each answer in the original equation. Only one of the answers worked, so that is our real solution.

🎯 Exam Tip: Always verify solutions by substituting them back into the original equation when you square both sides. This step is essential to identify and discard any extraneous solutions that may arise.

Question 12. \( 2\sqrt{2x+1} - 2x = 1 \)
Answer:
We have the equation: \( 2\sqrt{2x+1} - 2x = 1 \)
First, isolate the square root term on one side of the equation:

\( \implies 2\sqrt{2x+1} = 1 + 2x \)
To eliminate the square root, square both sides of the equation:

\( \implies (2\sqrt{2x+1})^2 = (1 + 2x)^2 \)
Expand both sides:

\( \implies 4(2x + 1) = 1^2 + 2(1)(2x) + (2x)^2 \)

\( \implies 8x + 4 = 1 + 4x + 4x^2 \)
Rearrange into a standard quadratic equation by moving all terms to one side:

\( \implies 0 = 4x^2 + 4x - 8x + 1 - 4 \)

\( \implies 4x^2 - 4x - 3 = 0 \)
We need to factor this quadratic expression. We look for two numbers that multiply to \( 4 \times (-3) = -12 \) and add up to -4.

\( \implies 4x^2 - 6x + 2x - 3 = 0 \)
Now, factor by grouping:

\( \implies 2x(2x - 3) + 1(2x - 3) = 0 \)

\( \implies (2x - 3)(2x + 1) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( 2x - 3 = 0 \), which means \( 2x = 3 \)
\( \implies x = \frac{3}{2} \)
Or \( 2x + 1 = 0 \), which means \( 2x = -1 \)
\( \implies x = -\frac{1}{2} \)

**Check the solutions:** It is crucial to check these potential solutions in the original equation, as squaring can introduce extraneous solutions.

Check for \( x = \frac{3}{2} \):
L.H.S. \( = 2\sqrt{2\left(\frac{3}{2}\right)+1} - 2\left(\frac{3}{2}\right) \)
\( = 2\sqrt{3+1} - 3 \)
\( = 2\sqrt{4} - 3 \)
\( = 2(2) - 3 = 4 - 3 = 1 \)
R.H.S. \( = 1 \)
Since L.H.S. = R.H.S., \( x = \frac{3}{2} \) is a valid solution.

Check for \( x = -\frac{1}{2} \):
L.H.S. \( = 2\sqrt{2\left(-\frac{1}{2}\right)+1} - 2\left(-\frac{1}{2}\right) \)
\( = 2\sqrt{-1+1} - (-1) \)
\( = 2\sqrt{0} + 1 \)
\( = 2(0) + 1 = 1 \)
R.H.S. \( = 1 \)
Since L.H.S. = R.H.S., \( x = -\frac{1}{2} \) is also a valid solution.
So, the valid solutions for \( x \) are \( \frac{3}{2} \) and \( -\frac{1}{2} \).
In simple words: We moved the non-square root part to one side and then squared both sides to remove the root. This led to a quadratic equation. After solving the quadratic equation for two possible \( x \) values, we checked both in the very first equation to make sure they were correct. In this case, both answers worked.

🎯 Exam Tip: Always isolate the square root term before squaring both sides of an equation to avoid complex cross-terms. Thoroughly check all potential solutions in the original equation to exclude extraneous solutions.

Question 13. \( \sqrt{4x-3}+\sqrt{2x+3} = 6 \)
Answer:
We have the equation: \( \sqrt{4x-3}+\sqrt{2x+3} = 6 \)
To solve this, we need to isolate one square root term and then square both sides.
Move one square root term to the right side:

\( \implies \sqrt{4x-3} = 6 - \sqrt{2x+3} \)
Now, square both sides to eliminate one square root:

\( \implies (\sqrt{4x-3})^2 = (6 - \sqrt{2x+3})^2 \)
Expand the right side using \( (a-b)^2 = a^2 - 2ab + b^2 \).

\( \implies 4x - 3 = 6^2 - 2(6)\sqrt{2x+3} + (\sqrt{2x+3})^2 \)

\( \implies 4x - 3 = 36 - 12\sqrt{2x+3} + (2x + 3) \)
Combine constant and \( x \) terms on the right side:

\( \implies 4x - 3 = 39 + 2x - 12\sqrt{2x+3} \)
Now, isolate the remaining square root term:

\( \implies 4x - 2x - 3 - 39 = -12\sqrt{2x+3} \)

\( \implies 2x - 42 = -12\sqrt{2x+3} \)
Divide the entire equation by 2 to simplify:

\( \implies x - 21 = -6\sqrt{2x+3} \)
To eliminate the final square root, square both sides again:

\( \implies (x - 21)^2 = (-6\sqrt{2x+3})^2 \)
Expand both sides:

\( \implies x^2 - 2(x)(21) + 21^2 = (-6)^2 (\sqrt{2x+3})^2 \)

\( \implies x^2 - 42x + 441 = 36(2x + 3) \)
Distribute on the right side:

\( \implies x^2 - 42x + 441 = 72x + 108 \)
Rearrange into a standard quadratic equation:

\( \implies x^2 - 42x - 72x + 441 - 108 = 0 \)

\( \implies x^2 - 114x + 333 = 0 \)
We need to factor this quadratic expression. We look for two numbers that multiply to 333 and add up to -114. These numbers are -111 and -3.

\( \implies x^2 - 111x - 3x + 333 = 0 \)
Now, factor by grouping:

\( \implies x(x - 111) - 3(x - 111) = 0 \)

\( \implies (x - 111)(x - 3) = 0 \)
This means either the first factor is zero or the second factor is zero.

Either \( x - 111 = 0 \), which means \( x = 111 \)
Or \( x - 3 = 0 \), which means \( x = 3 \)

**Check the solutions:** We must check these potential solutions in the *original* equation to ensure they are valid, as we squared twice.

Check for \( x = 111 \):
L.H.S. \( = \sqrt{4(111)-3} + \sqrt{2(111)+3} \)
\( = \sqrt{444-3} + \sqrt{222+3} \)
\( = \sqrt{441} + \sqrt{225} \)
\( = 21 + 15 = 36 \)
R.H.S. \( = 6 \)
Since L.H.S. \( \neq \) R.H.S. (36 is not equal to 6), \( x = 111 \) is an extraneous solution.

Check for \( x = 3 \):
L.H.S. \( = \sqrt{4(3)-3} + \sqrt{2(3)+3} \)
\( = \sqrt{12-3} + \sqrt{6+3} \)
\( = \sqrt{9} + \sqrt{9} \)
\( = 3 + 3 = 6 \)
R.H.S. \( = 6 \)
Since L.H.S. = R.H.S., \( x = 3 \) is a valid solution.
Therefore, the only valid solution is \( x = 3 \).
In simple words: When there are two square roots, we move one to the other side and square both sides once. This helps remove one root, but usually leaves another one. We then isolate the remaining square root and square both sides again to get a simple quadratic equation. After finding two possible answers, we carefully checked both in the very first equation to find out which one was truly correct. Only \( x=3 \) worked.

🎯 Exam Tip: For equations with multiple square roots, isolate one root and square, then isolate the remaining root and square again. Always perform a final check of all potential solutions in the original equation, as multiple squaring operations often introduce extraneous solutions.