CBSE Class 10 Mathematics Quadratic Equations Assignment Set 14

Solve the following equations for x:

Question 1. \( \frac{x}{x+1} + \frac{x+1}{x} = \frac{34}{15} \) , \( x \neq 0, -1 \)
Answer: Let us substitute \( y = \frac{x}{x+1} \). Then, the given equation can be written as:
\( y + \frac{1}{y} = \frac{34}{15} \)
\( \implies \frac{y^2 + 1}{y} = \frac{34}{15} \)
\( \implies 15(y^2 + 1) = 34y \)
\( \implies 15y^2 - 34y + 15 = 0 \)
Now, we factor the quadratic equation by splitting the middle term:
\( 15y^2 - 25y - 9y + 15 = 0 \)
\( \implies 5y(3y - 5) - 3(3y - 5) = 0 \)
\( \implies (5y - 3)(3y - 5) = 0 \)
This gives us two possible values for \( y \):
\( y = \frac{3}{5} \) or \( y = \frac{5}{3} \)
Now, we substitute back \( y = \frac{x}{x+1} \) to solve for \( x \):
When \( y = \frac{3}{5} \):
\( \frac{x}{x+1} = \frac{3}{5} \)
\( \implies 5x = 3(x + 1) \)
\( \implies 5x = 3x + 3 \)
\( \implies 2x = 3 \)
\( \implies x = \frac{3}{2} \)
When \( y = \frac{5}{3} \):
\( \frac{x}{x+1} = \frac{5}{3} \)
\( \implies 3x = 5(x + 1) \)
\( \implies 3x = 5x + 5 \)
\( \implies -2x = 5 \)
\( \implies x = -\frac{5}{2} \)
Thus, the solutions are \( x = \frac{3}{2} \) and \( x = -\frac{5}{2} \).
In simple words: To make the equation easier to solve, we temporarily replace \( \frac{x}{x+1} \) with \( y \). This gives a standard quadratic equation, which we solve for \( y \) first, and then we use those values to find \( x \).

Exam Tip: Substituting a complex term with a single variable like \( y \) simplifies the equation dramatically and helps avoid algebraic mistakes.

Question 2. \( \left(\frac{x-2}{x+2}\right)^2 + 3 - 4\left(\frac{x-2}{x+2}\right) = 0 \) , \( x \neq -2 \)
Answer: Let \( y = \frac{x-2}{x+2} \). Substituting this into the equation gives:
\( y^2 - 4y + 3 = 0 \)
Solving this quadratic equation by factoring:
\( y^2 - 3y - y + 3 = 0 \)
\( \implies y(y - 3) - 1(y - 3) = 0 \)
\( \implies (y - 1)(y - 3) = 0 \)
So, \( y = 1 \) or \( y = 3 \).
Case 1: When \( y = 1 \):
\( \frac{x-2}{x+2} = 1 \)
\( \implies x - 2 = x + 2 \)
\( \implies -2 = 2 \), which is not possible. Thus, \( y = 1 \) yields no real solution.
Case 2: When \( y = 3 \):
\( \frac{x-2}{x+2} = 3 \)
\( \implies x - 2 = 3(x + 2) \)
\( \implies x - 2 = 3x + 6 \)
\( \implies -2 - 6 = 3x - x \)
\( \implies -8 = 2x \)
\( \implies x = -4 \)
Therefore, the only valid solution is \( x = -4 \).
In simple words: Replacing the repeating fraction with \( y \) gives us two possible values: 1 and 3. Testing 1 gives an impossible statement, while testing 3 gives the final answer of -4.

Exam Tip: Always check if the obtained values for \( y \) produce logical equations for \( x \). If an equation simplifies to an impossible statement like \( -2 = 2 \), discard that case.

Question 3. \( (x^2-5x)^2 - 7(x^2-5x) + 6 = 0 \) , \( x \in \mathbb{R} \)
Answer: Let us substitute \( y = x^2 - 5x \). The equation becomes:
\( y^2 - 7y + 6 = 0 \)
Factorizing the equation:
\( y^2 - 6y - y + 6 = 0 \)
\( \implies y(y - 6) - 1(y - 6) = 0 \)
\( \implies (y - 1)(y - 6) = 0 \)
This gives \( y = 1 \) or \( y = 6 \).
Case 1: When \( y = 6 \):
\( x^2 - 5x = 6 \)
\( \implies x^2 - 5x - 6 = 0 \)
\( \implies x^2 - 6x + x - 6 = 0 \)
\( \implies x(x - 6) + 1(x - 6) = 0 \)
\( \implies (x - 6)(x + 1) = 0 \)
So, \( x = 6 \) or \( x = -1 \).
Case 2: When \( y = 1 \):
\( x^2 - 5x = 1 \)
\( \implies x^2 - 5x - 1 = 0 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-1)}}{2(1)} \)
\( \implies x = \frac{5 \pm \sqrt{25 + 4}}{2} \)
\( \implies x = \frac{5 \pm \sqrt{29}}{2} \)
Thus, the solutions are \( x = 6, -1, \frac{5 \pm \sqrt{29}}{2} \).
In simple words: Replacing \( x^2 - 5x \) with \( y \) lets us solve a simpler quadratic equation first. After finding \( y = 1 \) or \( y = 6 \), we solve the two resulting quadratic equations for \( x \).

Exam Tip: Remember to use the quadratic formula when a quadratic equation cannot be factored easily by splitting the middle term.

Question 4. \( \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}} = 2\frac{1}{6} \) , \( x \neq 0, 1 \)
Answer: Let \( y = \sqrt{\frac{x}{1-x}} \). Since \( \sqrt{\frac{1-x}{x}} \) is the reciprocal of \( y \), the equation can be written as:
\( y + \frac{1}{y} = \frac{13}{6} \)
\( \implies \frac{y^2 + 1}{y} = \frac{13}{6} \)
\( \implies 6(y^2 + 1) = 13y \)
\( \implies 6y^2 - 13y + 6 = 0 \)
We factor the quadratic equation by splitting the middle term:
\( 6y^2 - 9y - 4y + 6 = 0 \)
\( \implies 3y(2y - 3) - 2(2y - 3) = 0 \)
\( \implies (3y - 2)(2y - 3) = 0 \)
This gives us two cases for \( y \):
\( y = \frac{2}{3} \) or \( y = \frac{3}{2} \)
Now, we square both sides of our substitution to solve for \( x \):
Case 1: When \( y = \frac{2}{3} \):
\( \sqrt{\frac{x}{1-x}} = \frac{2}{3} \)
Squaring both sides:
\( \frac{x}{1-x} = \frac{4}{9} \)
\( \implies 9x = 4(1 - x) \)
\( \implies 9x = 4 - 4x \)
\( \implies 13x = 4 \)
\( \implies x = \frac{4}{13} \)
Case 2: When \( y = \frac{3}{2} \):
\( \sqrt{\frac{x}{1-x}} = \frac{3}{2} \)
Squaring both sides:
\( \frac{x}{1-x} = \frac{9}{4} \)
\( \implies 4x = 9(1 - x) \)
\( \implies 4x = 9 - 9x \)
\( \implies 13x = 9 \)
\( \implies x = \frac{9}{13} \)
Thus, the solutions are \( x = \frac{4}{13} \) and \( x = \frac{9}{13} \).
In simple words: We substitute \( y \) for the first square root term, making the second one \( \frac{1}{y} \). This simplifies the problem into a basic quadratic equation, which we solve and substitute back to get the final values.

Exam Tip: Be careful when squaring fractions like \( \frac{2}{3} \) and \( \frac{3}{2} \). A common mistake is forgetting to square both the numerator and the denominator.

Question 5. \( \sqrt{\frac{x^2+2}{x^2-2}} + 6\sqrt{\frac{x^2-2}{x^2+2}} = 5 \) , \( x \in \mathbb{R} \)
Answer: Let \( y = \sqrt{\frac{x^2+2}{x^2-2}} \). Substituting this gives the following equation:
\( y + \frac{6}{y} = 5 \)
\( \implies y^2 + 6 = 5y \)
\( \implies y^2 - 5y + 6 = 0 \)
Factorizing the equation:
\( y^2 - 2y - 3y + 6 = 0 \)
\( \implies y(y - 2) - 3(y - 2) = 0 \)
\( \implies (y - 2)(y - 3) = 0 \)
Thus, \( y = 2 \) or \( y = 3 \).
Case 1: When \( y = 2 \):
\( \sqrt{\frac{x^2+2}{x^2-2}} = 2 \)
Squaring both sides:
\( \frac{x^2+2}{x^2-2} = 4 \)
\( \implies x^2 + 2 = 4(x^2 - 2) \)
\( \implies x^2 + 2 = 4x^2 - 8 \)
\( \implies 3x^2 = 10 \)
\( \implies x^2 = \frac{10}{3} \)
\( \implies x = \pm\sqrt{\frac{10}{3}} \)
Case 2: When \( y = 3 \):
\( \sqrt{\frac{x^2+2}{x^2-2}} = 3 \)
Squaring both sides:
\( \frac{x^2+2}{x^2-2} = 9 \)
\( \implies x^2 + 2 = 9(x^2 - 2) \)
\( \implies x^2 + 2 = 9x^2 - 18 \)
\( \implies 8x^2 = 20 \)
\( \implies x^2 = \frac{20}{8} = \frac{5}{2} \)
\( \implies x = \pm\sqrt{\frac{5}{2}} \)
So, the solutions are \( x = \pm\sqrt{\frac{5}{2}} \) and \( x = \pm\sqrt{\frac{10}{3}} \).
In simple words: Replacing the square root term with \( y \) transforms the equation into a simple quadratic expression. Solving this for \( y \) gives us two values, which we square and solve to find the final four possible values for \( x \).

Exam Tip: Remember that taking the square root of a positive real number gives both positive and negative values, so do not forget to include the \( \pm \) symbol.

Question 6. \( \left(\frac{2x+1}{x-1}\right)^4 - 10\left(\frac{2x+1}{x-1}\right)^2 + 9 = 0 \) , \( x \in \mathbb{R} \)
Answer: Let us define \( y = \left(\frac{2x+1}{x-1}\right)^2 \). This simplifies the fourth-degree equation into a quadratic one:
\( y^2 - 10y + 9 = 0 \)
Factoring this equation:
\( y^2 - 9y - y + 9 = 0 \)
\( \implies y(y - 9) - 1(y - 9) = 0 \)
\( \implies (y - 1)(y - 9) = 0 \)
This gives \( y = 1 \) or \( y = 9 \).
Case 1: When \( y = 1 \):
\( \left(\frac{2x+1}{x-1}\right)^2 = 1 \)
Taking square roots on both sides:
\( \frac{2x+1}{x-1} = \pm 1 \)
Sub-case 1(a): \( \frac{2x+1}{x-1} = 1 \)
\( \implies 2x + 1 = x - 1 \)
\( \implies x = -2 \)
Sub-case 1(b): \( \frac{2x+1}{x-1} = -1 \)
\( \implies 2x + 1 = -(x - 1) \)
\( \implies 2x + 1 = -x + 1 \)
\( \implies 3x = 0 \)
\( \implies x = 0 \)
Case 2: When \( y = 9 \):
\( \left(\frac{2x+1}{x-1}\right)^2 = 9 \)
Taking square roots on both sides:
\( \frac{2x+1}{x-1} = \pm 3 \)
Sub-case 2(a): \( \frac{2x+1}{x-1} = 3 \)
\( \implies 2x + 1 = 3(x - 1) \)
\( \implies 2x + 1 = 3x - 3 \)
\( \implies x = 4 \)
Sub-case 2(b): \( \frac{2x+1}{x-1} = -3 \)
\( \implies 2x + 1 = -3(x - 1) \)
\( \implies 2x + 1 = -3x + 3 \)
\( \implies 5x = 2 \)
\( \implies x = \frac{2}{5} \)
Thus, the four solutions are \( x = -2, 0, 4, \frac{2}{5} \).
In simple words: We reduce the complex power-of-four equation to a quadratic equation by substituting \( y \). Solving for \( y \) gives two values, which lead to four linear equations for \( x \) after taking square roots.

Exam Tip: Always remember to take both positive and negative roots (e.g., \( \pm 1 \) and \( \pm 3 \)) when taking the square root of both sides, to ensure you don't miss any valid solutions.

Question 7. \( 2\left(x^2 + \frac{1}{x^2}\right) - 3\left(x + \frac{1}{x}\right) - 1 = 0 \) , \( x \in \mathbb{R} \)
Answer: Let \( y = x + \frac{1}{x} \).
Squaring both sides:
\( y^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 \)
\( \implies x^2 + \frac{1}{x^2} = y^2 - 2 \)
Now, substitute these expressions into the given equation:
\( 2(y^2 - 2) - 3y - 1 = 0 \)
\( \implies 2y^2 - 4 - 3y - 1 = 0 \)
\( \implies 2y^2 - 3y - 5 = 0 \)
Factorizing this quadratic equation:
\( 2y^2 - 5y + 2y - 5 = 0 \)
\( \implies y(2y - 5) + 1(2y - 5) = 0 \)
\( \implies (y + 1)(2y - 5) = 0 \)
This gives us \( y = -1 \) or \( y = \frac{5}{2} \).
Case 1: When \( y = -1 \):
\( x + \frac{1}{x} = -1 \)
\( \implies x^2 + x + 1 = 0 \)
The discriminant for this equation is \( D = b^2 - 4ac = 1^2 - 4(1)(1) = -3 \). Since the discriminant is less than 0, there are no real solutions for this case.
Case 2: When \( y = \frac{5}{2} \):
\( x + \frac{1}{x} = \frac{5}{2} \)
\( \implies \frac{x^2 + 1}{x} = \frac{5}{2} \)
\( \implies 2(x^2 + 1) = 5x \)
\( \implies 2x^2 - 5x + 2 = 0 \)
Factorizing:
\( 2x^2 - 4x - x + 2 = 0 \)
\( \implies 2x(x - 2) - 1(x - 2) = 0 \)
\( \implies (2x - 1)(x - 2) = 0 \)
Thus, \( x = 2 \) or \( x = \frac{1}{2} \).
Therefore, the real solutions are \( x = 2 \) and \( x = \frac{1}{2} \).
In simple words: By substituting \( y = x + \frac{1}{x} \), we can rewrite \( x^2 + \frac{1}{x^2} \) as \( y^2 - 2 \). This simplifies the problem into a quadratic equation in terms of \( y \). One value of \( y \) gives no real results, while the other gives two real numbers.

Exam Tip: Be sure to write down the check for the discriminant when a branch leads to complex roots, as proving why those roots are discarded is crucial for full credit.

Question 8. \( 6\left(x^2+\frac{1}{x^2}\right) - 25\left(x-\frac{1}{x}\right) + 12 = 0 \) , \( x \in \mathbb{R} \)
Answer: Let \( y = x - \frac{1}{x} \).
Squaring both sides:
\( y^2 = x^2 + \frac{1}{x^2} - 2 \)
\( \implies x^2 + \frac{1}{x^2} = y^2 + 2 \)
Now, substitute these into the original equation:
\( 6(y^2 + 2) - 25y + 12 = 0 \)
\( \implies 6y^2 + 12 - 25y + 12 = 0 \)
\( \implies 6y^2 - 25y + 24 = 0 \)
Factorizing by splitting the middle term (since \( 16 \times 9 = 144 \) and \( 16 + 9 = 25 \)):
\( 6y^2 - 16y - 9y + 24 = 0 \)
\( \implies 2y(3y - 8) - 3(3y - 8) = 0 \)
\( \implies (2y - 3)(3y - 8) = 0 \)
This gives two possible values: \( y = \frac{3}{2} \) or \( y = \frac{8}{3} \).
Case 1: When \( y = \frac{3}{2} \):
\( x - \frac{1}{x} = \frac{3}{2} \)
\( \implies \frac{x^2 - 1}{x} = \frac{3}{2} \)
\( \implies 2x^2 - 2 = 3x \)
\( \implies 2x^2 - 3x - 2 = 0 \)
Factorizing:
\( 2x^2 - 4x + x - 2 = 0 \)
\( \implies 2x(x - 2) + 1(x - 2) = 0 \)
\( \implies (2x + 1)(x - 2) = 0 \)
So, \( x = 2 \) or \( x = -\frac{1}{2} \).
Case 2: When \( y = \frac{8}{3} \):
\( x - \frac{1}{x} = \frac{8}{3} \)
\( \implies \frac{x^2 - 1}{x} = \frac{8}{3} \)
\( \implies 3x^2 - 3 = 8x \)
\( \implies 3x^2 - 8x - 3 = 0 \)
Factorizing:
\( 3x^2 - 9x + x - 3 = 0 \)
\( \implies 3x(x - 3) + 1(x - 3) = 0 \)
\( \implies (3x + 1)(x - 3) = 0 \)
So, \( x = 3 \) or \( x = -\frac{1}{3} \).
Thus, the four solutions are \( x = 3, -\frac{1}{3}, 2, -\frac{1}{2} \).
In simple words: Substituting \( y = x - \frac{1}{x} \) allows us to represent \( x^2 + \frac{1}{x^2} \) as \( y^2 + 2 \). This simplifies the algebraic equation into a basic quadratic, which yields two values of \( y \), and each of those gives two values of \( x \).

Exam Tip: Be careful with the sign in the algebraic identity: \( \left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 \) (uses a minus sign), which differs from the plus sign used in the previous question.

Question 9. \( 2x^4 - x^3 - 11x^2 - x + 2 = 0 \)
Answer: Since \( x = 0 \) does not satisfy the equation, we can divide the entire equation by \( x^2 \) to simplify it:
\( 2x^2 - x - 11 - \frac{1}{x} + \frac{2}{x^2} = 0 \)
Grouping the similar terms:
\( 2\left(x^2 + \frac{1}{x^2}\right) - \left(x + \frac{1}{x}\right) - 11 = 0 \)
Now, substitute \( y = x + \frac{1}{x} \), which gives \( x^2 + \frac{1}{x^2} = y^2 - 2 \):
\( 2(y^2 - 2) - y - 11 = 0 \)
\( \implies 2y^2 - 4 - y - 11 = 0 \)
\( \implies 2y^2 - y - 15 = 0 \)
Factorizing this quadratic equation:
\( 2y^2 - 6y + 5y - 15 = 0 \)
\( \implies 2y(y - 3) + 5(y - 3) = 0 \)
\( \implies (2y + 5)(y - 3) = 0 \)
This gives \( y = 3 \) or \( y = -\frac{5}{2} \).
Case 1: When \( y = 3 \):
\( x + \frac{1}{x} = 3 \)
\( \implies x^2 - 3x + 1 = 0 \)
Using the quadratic formula:
\( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)} \)
\( \implies x = \frac{3 \pm \sqrt{5}}{2} \)
Case 2: When \( y = -\frac{5}{2} \):
\( x + \frac{1}{x} = -\frac{5}{2} \)
\( \implies \frac{x^2 + 1}{x} = -\frac{5}{2} \)
\( \implies 2x^2 + 2 = -5x \)
\( \implies 2x^2 + 5x + 2 = 0 \)
Factorizing:
\( 2x^2 + 4x + x + 2 = 0 \)
\( \implies 2x(x + 2) + 1(x + 2) = 0 \)
\( \implies (2x + 1)(x + 2) = 0 \)
So, \( x = -2 \) or \( x = -\frac{1}{2} \).
Thus, the four solutions are \( x = -2, -\frac{1}{2}, \frac{3 \pm \sqrt{5}}{2} \).
In simple words: This is a symmetric equation, so dividing by \( x^2 \) and grouping the inverse terms allows us to use the substitution \( y = x + \frac{1}{x} \). This leads to two quadratic equations, which we solve for \( x \).

Exam Tip: A fourth-degree equation of symmetric form (where coefficients from the left and right match) can always be simplified by dividing by \( x^2 \) first.

Question 10. \( 6x^4 + 5x^3 - 38x^2 + 5x + 6 = 0 \)
Answer: Dividing the equation by \( x^2 \) since \( x \neq 0 \):
\( 6x^2 + 5x - 38 + \frac{5}{x} + \frac{6}{x^2} = 0 \)
Rearranging and grouping terms:
\( 6\left(x^2 + \frac{1}{x^2}\right) + 5\left(x + \frac{1}{x}\right) - 38 = 0 \)
Using the substitution \( y = x + \frac{1}{x} \) (which means \( x^2 + \frac{1}{x^2} = y^2 - 2 \)):
\( 6(y^2 - 2) + 5y - 38 = 0 \)
\( \implies 6y^2 - 12 + 5y - 38 = 0 \)
\( \implies 6y^2 + 5y - 50 = 0 \)
Factorizing by splitting the middle term (using \( 20 \) and \( -15 \prime \)):
\( 6y^2 + 20y - 15y - 50 = 0 \)
\( \implies 2y(3y + 10) - 5(3y + 10) = 0 \)
\( \implies (2y - 5)(3y + 10) = 0 \)
Thus, we have \( y = \frac{5}{2} \) or \( y = -\frac{10}{3} \).
Case 1: When \( y = \frac{5}{2} \):
\( x + \frac{1}{x} = \frac{5}{2} \)
\( \implies 2x^2 - 5x + 2 = 0 \)
\( \implies (2x - 1)(x - 2) = 0 \)
So, \( x = 2 \) or \( x = \frac{1}{2} \).
Case 2: When \( y = -\frac{10}{3} \):
\( x + \frac{1}{x} = -\frac{10}{3} \)
\( \implies 3x^2 + 10x + 3 = 0 \)
\( \implies (3x + 1)(x + 3) = 0 \)
So, \( x = -3 \) or \( x = -\frac{1}{3} \).
Thus, the four solutions are \( x = 2, \frac{1}{2}, -3, -\frac{1}{3} \).
In simple words: Dividing by \( x^2 \) helps group terms together to use substitution. Solving for \( y \) gives two values, which lead to four solutions for \( x \).

Exam Tip: Be methodical with factoring quadratic equations with large coefficients. Find factors of \( ac = -300 \) that sum to \( b = 5 \) carefully.

Question 11. \( 3^{2x-2} - 10 \cdot 3^{x-1} + 9 = 0 \)
Answer: Let us rewrite the given equation as:
\( \left(3^{x-1}\right)^2 - 10 \cdot 3^{x-1} + 9 = 0 \)
Now, substitute \( y = 3^{x-1} \). This gives us the quadratic equation:
\( y^2 - 10y + 9 = 0 \)
Factorizing the equation:
\( y^2 - 9y - y + 9 = 0 \)
\( \implies y(y - 9) - 1(y - 9) = 0 \)
\( \implies (y - 1)(y - 9) = 0 \)
This yields \( y = 1 \) or \( y = 9 \).
Now, we substitute back \( y = 3^{x-1} \) to solve for \( x \):
Case 1: When \( y = 1 \):
\( 3^{x-1} = 1 \)
\( \implies 3^{x-1} = 3^0 \)
\( \implies x - 1 = 0 \)
\( \implies x = 1 \)
Case 2: When \( y = 9 \):
\( 3^{x-1} = 9 \)
\( \implies 3^{x-1} = 3^2 \)
\( \implies x - 1 = 2 \)
\( \implies x = 3 \)
Thus, the solutions are \( x = 1 \) and \( x = 3 \).
In simple words: We replace \( 3^{x-1} \) with \( y \) to convert the expression into a standard quadratic equation. Solving for \( y \) gives 1 and 9, which we then convert back to find \( x = 1 \) and \( x = 3 \).

Exam Tip: Remember that any non-zero number raised to the power of 0 is 1. Thus, when solving \( 3^{x-1} = 1 \), equate the exponent to 0.

Question 12. \( x^{\frac{2}{3}} + x^{\frac{1}{3}} = 2 \)
Answer: Let us substitute \( y = x^{\frac{1}{3}} \). Then \( x^{\frac{2}{3}} = y^2 \). Substituting this into our equation gives:
\( y^2 + y = 2 \)
\( \implies y^2 + y - 2 = 0 \)
Factorizing by splitting the middle term:
\( y^2 + 2y - y - 2 = 0 \)
\( \implies y(y + 2) - 1(y + 2) = 0 \)
\( \implies (y - 1)(y + 2) = 0 \)
This gives \( y = 1 \) or \( y = -2 \).
Now, we substitute back \( y = x^{\frac{1}{3}} \) to find \( x \):
Case 1: When \( y = 1 \):
\( x^{\frac{1}{3}} = 1 \)
Cubing both sides:
\( x = 1^3 = 1 \)
Case 2: When \( y = -2 \):
\( x^{\frac{1}{3}} = -2 \)
Cubing both sides:
\( x = (-2)^3 = -8 \)
Thus, the solutions are \( x = 1 \) and \( x = -8 \).
In simple words: Replacing \( x^{\frac{1}{3}} \) with \( y \) gives us a basic quadratic equation. Solving for \( y \) gives 1 and -2, and cubing these values gives the final answers for \( x \).

Exam Tip: When cubing a negative number, the result remains negative: \( (-2)^3 = -8 \). This is different from squaring, where the result is always non-negative.

Question 13. \( \frac{\sqrt{x}}{2} + \frac{16}{x\sqrt{x}} = \frac{6}{\sqrt{x}} \)
Answer: To eliminate the fractional terms, let us multiply the entire equation by the common denominator, which is \( 2x\sqrt{x} \) (assuming \( x > 0 \)):
\( 2x\sqrt{x} \left( \frac{\sqrt{x}}{2} \right) + 2x\sqrt{x} \left( \frac{16}{x\sqrt{x}} \right) = 2x\sqrt{x} \left( \frac{6}{\sqrt{x}} \right) \)
Simplifying each term:
\( x^2 + 32 = 12x \)
\( \implies x^2 - 12x + 32 = 0 \)
Factorizing the quadratic equation:
\( x^2 - 8x - 4x + 32 = 0 \)
\( \implies x(x - 8) - 4(x - 8) = 0 \)
\( \implies (x - 4)(x - 8) = 0 \)
This gives us two solutions:
\( x = 4 \) or \( x = 8 \).
Since both values are positive, they satisfy the initial condition of \( x > 0 \) for the square roots to be real. Thus, the solutions are \( x = 4 \) and \( x = 8 \).
In simple words: Multiplying everything by \( 2x\sqrt{x} \) clears out the fractions and square roots, turning the problem into a standard quadratic equation that we can easily factor to get 4 and 8.

Exam Tip: Whenever you clear denominators containing variables, always verify that the final solutions do not make any original denominator equal to zero.

Question 14. \( 3^{x+2} + 3^{-x} = 10 \)
Answer: Using properties of exponents, we can rewrite the equation as:
\( 3^2 \cdot 3^x + \frac{1}{3^x} = 10 \)
\( \implies 9 \cdot 3^x + \frac{1}{3^x} = 10 \)
Let us substitute \( y = 3^x \). The equation becomes:
\( 9y + \frac{1}{y} = 10 \)
Multiplying by \( y \) on both sides:
\( 9y^2 + 1 = 10y \)
\( \implies 9y^2 - 10y + 1 = 0 \)
Factorizing the equation:
\( 9y^2 - 9y - y + 1 = 0 \)
\( \implies 9y(y - 1) - 1(y - 1) = 0 \)
\( \implies (9y - 1)(y - 1) = 0 \)
This gives \( y = \frac{1}{9} \) or \( y = 1 \).
Substitute back \( y = 3^x \):
Case 1: When \( y = \frac{1}{9} \):
\( 3^x = \frac{1}{9} = 3^{-2} \)
\( \implies x = -2 \)
Case 2: When \( y = 1 \):
\( 3^x = 1 = 3^0 \)
\( \implies x = 0 \)
Thus, the solutions are \( x = 0 \) and \( x = -2 \).
In simple words: We rewrite \( 3^{-x} \) as \( \frac{1}{3^x} \) and replace \( 3^x \) with \( y \). This creates a simple quadratic equation. Solving for \( y \) gives two fractions, which correspond to \( x = 0 \) and \( x = -2 \).

Exam Tip: Be comfortable with rewriting negative exponents as fractions, i.e., \( a^{-n} = \frac{1}{a^n} \), which is a key step in simplifying exponential equations.

Question 15. \( (a+x)^{\frac{1}{3}} + (b-x)^{\frac{1}{3}} = (a+b)^{\frac{1}{3}} \)
Answer: Let us cube both sides of the equation using the identity \( (u + v)^3 = u^3 + v^3 + 3uv(u + v) \):
Let \( u = (a+x)^{\frac{1}{3}} \) and \( v = (b-x)^{\frac{1}{3}} \). The equation is \( u + v = (a+b)^{\frac{1}{3}} \).
Cubing both sides:
\( (u + v)^3 = \left((a+b)^{\frac{1}{3}}\right)^3 \)
\( \implies u^3 + v^3 + 3uv(u + v) = a+b \)
Substitute back the definitions of \( u \) and \( v \), noting that \( u^3 = a+x \) and \( v^3 = b-x \):
\( (a + x) + (b - x) + 3(a+x)^{\frac{1}{3}}(b-x)^{\frac{1}{3}}\left((a+x)^{\frac{1}{3}} + (b-x)^{\frac{1}{3}}\right) = a+b \)
\( \implies a + b + 3(a+x)^{\frac{1}{3}}(b-x)^{\frac{1}{3}}(a+b)^{\frac{1}{3}} = a+b \)
Subtracting \( a + b \) from both sides:
\( 3(a+x)^{\frac{1}{3}}(b-x)^{\frac{1}{3}}(a+b)^{\frac{1}{3}} = 0 \)
Assuming \( a+b \neq 0 \), we can divide by \( 3(a+b)^{\frac{1}{3}} \):
\( (a+x)^{\frac{1}{3}}(b-x)^{\frac{1}{3}} = 0 \)
This gives two cases:
Case 1: \( (a+x)^{\frac{1}{3}} = 0 \)
\( \implies a + x = 0 \)
\( \implies x = -a \)
Case 2: \( (b-x)^{\frac{1}{3}} = 0 \)
\( \implies b - x = 0 \)
\( \implies x = b \)
Thus, the solutions are \( x = -a \) and \( x = b \).
In simple words: Cubing both sides and applying the cubic expansion formula simplifies the equation by cancelling out several terms, eventually leaving us with two simple factors that give the roots \( -a \) and \( b \).

Exam Tip: Using the algebraic identity \( (u+v)^3 = u^3 + v^3 + 3uv(u+v) \) instead of the standard expanded form is extremely helpful here, as it allows direct substitution of the original equation.

Question 16. \( 9^{x+2} - 6 \cdot 3^{x+1} + 1 = 0 \)
Answer: We can rewrite the bases of the terms using powers of 3:
\( 9^{x+2} = \left(3^2\right)^{x+2} = 3^{2x+4} = 3^4 \cdot 3^{2x} = 81 \cdot \left(3^x\right)^2 \)
\( 6 \cdot 3^{x+1} = 6 \cdot 3 \cdot 3^x = 18 \cdot 3^x \)
Substituting these back into the original equation:
\( 81\left(3^x\right)^2 - 18\cdot 3^x + 1 = 0 \)
Now, let us substitute \( y = 3^x \):
\( 81y^2 - 18y + 1 = 0 \)
This is a perfect square trinomial:
\( (9y - 1)^2 = 0 \)
\( \implies 9y - 1 = 0 \)
\( \implies y = \frac{1}{9} \)
Now, substitute back \( y = 3^x \):
\( 3^x = \frac{1}{9} = 3^{-2} \)
\( \implies x = -2 \)
Thus, the solution is \( x = -2 \).
In simple words: Expressing everything with a base of 3 turns this into a perfect square quadratic equation in terms of \( 3^x \). Solving it gives \( 3^x = 1/9 \), which means \( x = -2 \).

Exam Tip: Recognizing perfect squares (like \( 81y^2 - 18y + 1 = (9y-1)^2 \)) saves valuable time during exams and prevents factoring errors.

Question 17. \( 2^{x+3} + 2^{-x} = 6 \)
Answer: Using exponential rules, we can rewrite the equation as:
\( 2^3 \cdot 2^x + \frac{1}{2^x} = 6 \)
\( \implies 8 \cdot 2^x + \frac{1}{2^x} = 6 \)
Substitute \( y = 2^x \):
\( 8y + \frac{1}{y} = 6 \)
Multiplying by \( y \) on both sides:
\( 8y^2 + 1 = 6y \)
\( \implies 8y^2 - 6y + 1 = 0 \)
Factorizing the quadratic equation:
\( 8y^2 - 4y - 2y + 1 = 0 \)
\( \implies 4y(2y - 1) - 1(2y - 1) = 0 \)
\( \implies (4y - 1)(2y - 1) = 0 \)
This gives us \( y = \frac{1}{4} \) or \( y = \frac{1}{2} \).
Substitute back \( y = 2^x \):
Case 1: When \( y = \frac{1}{4} \):
\( 2^x = \frac{1}{4} = 2^{-2} \)
\( \implies x = -2 \)
Case 2: When \( y = \frac{1}{2} \):
\( 2^x = \frac{1}{2} = 2^{-1} \)
\( \implies x = -1 \)
Thus, the solutions are \( x = -1 \) and \( x = -2 \).
In simple words: We convert \( 2^{-x} \) to a fraction and substitute \( y \) for \( 2^x \). This yields a quadratic equation whose solutions are fractions, giving us \( x = -1 \) and \( x = -2 \).

Exam Tip: Express fractions like \( 1/4 \) and \( 1/2 \) as powers of 2 (i.e., \( 2^{-2} \) and \( 2^{-1} \)) to directly find the exponent values.

Question 18. \( 5^{1+x} + 5^{1-x} = 26 \)
Answer: Rewrite the terms using the rules of exponents:
\( 5 \cdot 5^x + \frac{5}{5^x} = 26 \)
Now, substitute \( y = 5^x \):
\( 5y + \frac{5}{y} = 26 \)
Multiplying the entire equation by \( y \):
\( 5y^2 + 5 = 26y \)
\( \implies 5y^2 - 26y + 5 = 0 \)
Factorizing by splitting the middle term:
\( 5y^2 - 25y - y + 5 = 0 \)
\( \implies 5y(y - 5) - 1(y - 5) = 0 \)
\( \implies (5y - 1)(y - 5) = 0 \)
This yields \( y = \frac{1}{5} \) or \( y = 5 \).
Now substitute back \( y = 5^x \):
Case 1: When \( y = \frac{1}{5} \):
\( 5^x = \frac{1}{5} = 5^{-1} \)
\( \implies x = -1 \)
Case 2: When \( y = 5 \):
\( 5^x = 5 = 5^1 \)
\( \implies x = 1 \)
Thus, the solutions are \( x = -1 \) and \( x = 1 \).
In simple words: Replacing \( 5^x \) with \( y \) converts the exponent problem into a basic quadratic equation. Solving for \( y \) gives \( 1/5 \) and \( 5 \), which translates to \( x = -1 \) and \( x = 1 \).

Exam Tip: Be careful not to lose the constant factor when rewriting \( 5^{1-x} \) as \( \frac{5}{5^x} \). Both terms on the left contain a factor of 5.

Question 19. \( 5^{2x} - 5^{x+3} + 125 = 5^x \)
Answer: Let us simplify the exponent in the second term:
\( 5^{x+3} = 5^3 \cdot 5^x = 125 \cdot 5^x \)
Substituting this back into the equation:
\( 5^{2x} - 125 \cdot 5^x + 125 = 5^x \)
Move all terms to the left side:
\( 5^{2x} - 125 \cdot 5^x - 5^x + 125 = 0 \)
\( \implies \left(5^x\right)^2 - 126 \cdot 5^x + 125 = 0 \)
Now, substitute \( y = 5^x \):
\( y^2 - 126y + 125 = 0 \)
Factorizing the equation:
\( y^2 - 125y - y + 125 = 0 \)
\( \implies y(y - 125) - 1(y - 125) = 0 \)
\( \implies (y - 1)(y - 125) = 0 \)
This gives \( y = 1 \) or \( y = 125 \).
Substitute back \( y = 5^x \):
Case 1: When \( y = 1 \):
\( 5^x = 1 = 5^0 \)
\( \implies x = 0 \)
Case 2: When \( y = 125 \):
\( 5^x = 125 = 5^3 \)
\( \implies x = 3 \)
Thus, the solutions are \( x = 0 \) and \( x = 3 \).
In simple words: Grouping terms and substituting \( y = 5^x \) gives us a quadratic equation. Factoring it yields \( y = 1 \) and \( y = 125 \), which gives \( x = 0 \) and \( x = 3 \).

Exam Tip: Don't forget to combine like terms on one side (like \( -125 \cdot 5^x \) and \( -5^x \)) before substituting and trying to solve the quadratic equation.

Question 20. \( (x^2+3x+2)^2 - 8(x^2+3x) - 4 = 0 \)
Answer: Let us substitute \( y = x^2 + 3x \). The given equation can then be written as:
\( (y + 2)^2 - 8y - 4 = 0 \)
Expanding the squared term:
\( y^2 + 4y + 4 - 8y - 4 = 0 \)
\( \implies y^2 - 4y = 0 \)
\( \implies y(y - 4) = 0 \)
This gives us two possibilities:
\( y = 0 \) or \( y = 4 \).
Now, we substitute back \( y = x^2 + 3x \) to find \( x \):
Case 1: When \( y = 0 \):
\( x^2 + 3x = 0 \)
\( \implies x(x + 3) = 0 \)
So, \( x = 0 \) or \( x = -3 \).
Case 2: When \( y = 4 \):
\( x^2 + 3x = 4 \)
\( \implies x^2 + 3x - 4 = 0 \)
Factorizing:
\( x^2 + 4x - x - 4 = 0 \)
\( \implies x(x + 4) - 1(x + 4) = 0 \)
\( \implies (x - 1)(x + 4) = 0 \)
So, \( x = 1 \) or \( x = -4 \).
Therefore, the solutions are \( x = 0, 1, -3, -4 \).
In simple words: Replacing the common part \( x^2 + 3x \) with \( y \) simplifies the problem. We solve the resulting equation for \( y \), which gives 0 and 4, and then solve those two quadratic equations to find the final four values of \( x \).

Exam Tip: Expanding \( (y+2)^2 \) correctly as \( y^2 + 4y + 4 \) is critical. A common mistake is writing it as just \( y^2 + 4 \).

Question 21. \( (x^2+3x)^2 - (x^2+3x) - 6 = 0 \)
Answer: Let us substitute \( y = x^2 + 3x \). The equation simplifies to:
\( y^2 - y - 6 = 0 \)
Factorizing the quadratic equation:
\( y^2 - 3y + 2y - 6 = 0 \)
\( \implies y(y - 3) + 2(y - 3) = 0 \)
\( \implies (y - 3)(y + 2) = 0 \)
This gives \( y = 3 \) or \( y = -2 \).
Now, substitute back \( y = x^2 + 3x \):
Case 1: When \( y = 3 \):
\( x^2 + 3x = 3 \)
\( \implies x^2 + 3x - 3 = 0 \)
Using the quadratic formula:
\( x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-3)}}{2(1)} \)
\( \implies x = \frac{-3 \pm \sqrt{9 + 12}}{2} \)
\( \implies x = \frac{-3 \pm \sqrt{21}}{2} \)
Case 2: When \( y = -2 \):
\( x^2 + 3x = -2 \)
\( \implies x^2 + 3x + 2 = 0 \)
Factorizing:
\( x^2 + 2x + x + 2 = 0 \)
\( \implies x(x + 2) + 1(x + 2) = 0 \)
\( \implies (x + 1)(x + 2) = 0 \)
So, \( x = -1 \) or \( x = -2 \).
Thus, the four solutions are \( x = -2, -1, \frac{-3 \pm \sqrt{21}}{2} \).
In simple words: Replacing \( x^2+3x \) with \( y \) allows us to solve a simpler quadratic equation. This gives us two values for \( y \), which we substitute back to get the final four values for \( x \).

Exam Tip: Since \( \sqrt{21} \) cannot be simplified further, leave the roots of that branch in fraction form as \( \frac{-3 \pm \sqrt{21}}{2} \).

Question 22. \( (x^2-2x)^2 - 4(x^2-2x) + 3 = 0 \)
Answer: Let \( y = x^2 - 2x \). The equation becomes:
\( y^2 - 4y + 3 = 0 \)
Factorizing the equation:
\( y^2 - 3y - y + 3 = 0 \)
\( \implies y(y - 3) - 1(y - 3) = 0 \)
\( \implies (y - 1)(y - 3) = 0 \)
This gives us \( y = 1 \) or \( y = 3 \).
Substitute back \( y = x^2 - 2x \):
Case 1: When \( y = 3 \):
\( x^2 - 2x = 3 \)
\( \implies x^2 - 2x - 3 = 0 \)
Factorizing:
\( x^2 - 3x + x - 3 = 0 \)
\( \implies x(x - 3) + 1(x - 3) = 0 \)
\( \implies (x + 1)(x - 3) = 0 \)
So, \( x = -1 \) or \( x = 3 \).
Case 2: When \( y = 1 \):
\( x^2 - 2x = 1 \)
\( \implies x^2 - 2x - 1 = 0 \)
Using the quadratic formula:
\( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} \)
\( \implies x = \frac{2 \pm \sqrt{4 + 4}}{2} \)
\( \implies x = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} \)
\( \implies x = 1 \pm \sqrt{2} \)
Thus, the four solutions are \( x = -1, 3, 1 \pm \sqrt{2} \).
In simple words: We replace \( x^2 - 2x \) with \( y \) to solve a simpler quadratic equation first. This gives us two quadratic equations in \( x \), which we solve to find the final four values.

Exam Tip: Remember to simplify the surd: \( \sqrt{8} = 2\sqrt{2} \), which allows you to cancel the common factor of 2 in the numerator and denominator.

Question 23. \( 9x^4 - 30x^2 + 25 = 0 \)
Answer: Let \( y = x^2 \). Then \( x^4 = y^2 \). Substituting this gives:
\( 9y^2 - 30y + 25 = 0 \)
Notice that this is a perfect square trinomial:
\( (3y - 5)^2 = 0 \)
\( \implies 3y - 5 = 0 \)
\( \implies y = \frac{5}{3} \)
Now, substitute back \( y = x^2 \):
\( x^2 = \frac{5}{3} \)
Taking the square root of both sides:
\( x = \pm\sqrt{\frac{5}{3}} \)
Thus, the solutions are \( x = \pm\sqrt{\frac{5}{3}} \).
In simple words: Substituting \( y = x^2 \) turns this into a perfect square quadratic. Solving for \( y \) gives \( 5/3 \), and taking the square root of that gives the final positive and negative solutions.

Exam Tip: If you don't recognize the perfect square immediately, you can still use the middle term splitting method with \( -15y \) and \( -15y \) to factor it.

Question 24. \( 25x^4 - 20x^2 + 4 = 0 \)
Answer: Let us substitute \( y = x^2 \). The equation becomes:
\( 25y^2 - 20y + 4 = 0 \)
This is a perfect square trinomial:
\( (5y - 2)^2 = 0 \)
\( \implies 5y - 2 = 0 \)
\( \implies y = \frac{2}{5} \)
Now substitute back \( y = x^2 \):
\( x^2 = \frac{2}{5} \)
Taking square roots on both sides:
\( x = \pm\sqrt{\frac{2}{5}} \)
Thus, the solutions are \( x = \pm\sqrt{\frac{2}{5}} \).
In simple words: This equation is a perfect square of \( 5x^2 - 2 \). Setting it to zero and solving for \( x \) gives two symmetric square root answers.

Exam Tip: Always make sure to write the \( \pm \) symbol when solving \( x^2 = \text{constant} \), as missing the negative solution is a common way to lose points.

Question 25. \( x^3 + 3x^2 - 4x - 12 = 0 \)
Answer: We can solve this cubic equation by grouping the terms:
\( \left(x^3 + 3x^2\right) - \left(4x + 12\right) = 0 \)
Take out the common factors from each group:
\( x^2(x + 3) - 4(x + 3) = 0 \)
Factor out the common binomial \( (x + 3) \):
\( (x^2 - 4)(x + 3) = 0 \)
We can factor the difference of squares \( x^2 - 4 \) as \( (x - 2)(x + 2) \):
\( (x - 2)(x + 2)(x + 3) = 0 \)
Setting each factor to zero:
\( x - 2 = 0 \implies x = 2 \)
\( x + 2 = 0 \implies x = -2 \)
\( x + 3 = 0 \implies x = -3 \)
Thus, the solutions are \( x = 2, -2, -3 \).
In simple words: By grouping the first two terms and the last two terms, we find a common factor of \( x+3 \). Factoring it out leaves us with a difference of squares that is easily broken down into simple roots.

Exam Tip: Look for opportunities to group terms in cubic polynomials. If the ratios of the coefficients of the first two terms and the last two terms are equal, grouping will always work.