Test your GMAT problem-solving skills with the following set of 10 quadratic equation problems. Detailed step-by-step solutions and explanations are integrated directly below each question.
Practice Questions with Solutions
Question 1: Solve the equation \(x^2 + 5x + 6 = 0\)
Options:
- \(x = 2\) and \(x = 3\)
- \(x = -2\) only
- \(x = -2\) and \(x = -3\)
- \(x = -3\) only
- No solutions
Solution & Explanation:
Factor the quadratic expression:
\[ (x + 2)(x + 3) = 0 \]Solve each factor:
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]Correct answer: C
Question 2: Find the values of constants \(k\) and \(m\) such that \(2x^2 + kx - m = 0\) has solutions \(x = 1\) and \(x = -2\).
Options:
- \(k = 1, m = -2\)
- \(k = -2, m = -4\)
- \(k = 2, m = 0\)
- \(k = 2, m = 4\)
- \(k = 0, m = 4\)
Solution & Explanation:
Substitute \(x = 1\):
\[ 2(1)^2 + k(1) - m = 0 \quad \Rightarrow \quad k - m = -2 \]Substitute \(x = -2\):
\[ 2(-2)^2 + k(-2) - m = 0 \quad \Rightarrow \quad -2k - m = -8 \]Solve the system of equations by subtracting the second equation from the first:
\[ (k - m) - (-2k - m) = -2 - (-8) \quad \Rightarrow \quad 3k = 6 \quad \Rightarrow \quad k = 2 \]Substitute \(k = 2\) back:
\[ 2 - m = -2 \quad \Rightarrow \quad m = 4 \]Correct answer: D
Question 3: Solve the equation \((x - 1)(x + 3) = 1 - x\)
Options:
- \(x = 1\) and \(x = -3\)
- \(x = 1\) and \(x = -4\)
- \(x = 1\) and \(x = 4\)
- \(x = 1\) and \(x = 0\)
- \(x = 1\) and \(x = 3\)
Solution & Explanation:
Bring all terms to one side:
\[ (x - 1)(x + 3) - (1 - x) = 0 \quad \Rightarrow \quad (x - 1)(x + 3) + (x - 1) = 0 \]Factor out \((x - 1)\):
\[ (x - 1)((x + 3) + 1) = (x - 1)(x + 4) = 0 \]Solve for \(x\):
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]Correct answer: B
Question 4: Solve: \(2 - (x - 2)^2 = -18\)
Options:
- \(x = 2 + 2\sqrt{5}, x = 2 - 2\sqrt{5}\)
- \(x = 2\sqrt{5}, x = -2\sqrt{5}\)
- \(x = 20, x = -20\)
- \(x = \sqrt{5}, x = -\sqrt{5}\)
- \(x = -2 + 2\sqrt{5}, x = -2 - 2\sqrt{5}\)
Solution & Explanation:
Subtract 2 from both sides:
\[ -(x - 2)^2 = -20 \]Multiply by \(-1\):
\[ (x - 2)^2 = 20 \]Take the square root of both sides:
\[ x - 2 = \pm \sqrt{20} = \pm 2\sqrt{5} \quad \Rightarrow \quad x = 2 \pm 2\sqrt{5} \]Correct answer: A
Question 5: Find the positive value of \(k\) such that the equation \(x^2 + kx + 4 = 0\) has exactly one solution.
Options:
- 4
- 3
- 2
- 1
- -2
Solution & Explanation:
For a quadratic equation to have exactly one solution, its discriminant must be zero (\(b^2 - 4ac = 0\)):
\[ k^2 - 4(1)(4) = 0 \quad \Rightarrow \quad k^2 - 16 = 0 \quad \Rightarrow \quad k = 4 \quad (\text{since } k > 0) \]Correct answer: A
Question 6: If \(x = 2\) is a solution of \((m+2)x^2 = 16\), find the second solution.
Options:
- \(x = 3\)
- \(x = -3\)
- \(x = -2\)
- \(x = -4\)
- \(x = 0\)
Solution & Explanation:
Substitute \(x = 2\):
\[ (m + 2)(2)^2 = 16 \quad \Rightarrow \quad 4(m + 2) = 16 \quad \Rightarrow \quad m + 2 = 4 \quad \Rightarrow \quad m = 2 \]Substitute \(m = 2\) back into the equation:
\[ 4x^2 = 16 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = \pm 2 \]The second solution is \(x = -2\).
Correct answer: C
Question 7: If \(x = -2\) is a solution of \(\dfrac{1}{2} a x^2 + 3x - 4 = 0\), find the second solution.
Options:
- \(x = \dfrac{4}{5}\)
- \(x = 4\)
- \(x = \dfrac{5}{4}\)
- \(x = 5\)
- No second solution
Solution & Explanation:
Substitute \(x = -2\):
\[ \frac{1}{2}a(-2)^2 + 3(-2) - 4 = 0 \quad \Rightarrow \quad 2a - 6 - 4 = 0 \quad \Rightarrow \quad 2a - 10 = 0 \quad \Rightarrow \quad a = 5 \]Substitute \(a = 5\):
\[ \frac{5}{2}x^2 + 3x - 4 = 0 \quad \Rightarrow \quad (x + 2)\left(\frac{5}{2}x - 2\right) = 0 \]Solving the second factor:
\[ \frac{5}{2}x - 2 = 0 \quad \Rightarrow \quad x = \frac{4}{5} \]Correct answer: A
Question 8: Find \(b\) and \(c\) so that \(x^2 + bx + c = 0\) has solutions \(x = 2\) and \(x = -1\).
Options:
- \(b = 2, c = -1\)
- \(b = -2, c = 1\)
- \(b = 2, c = 2\)
- \(b = 1, c = 0\)
- \(b = -1, c = -2\)
Solution & Explanation:
Using the roots, the quadratic can be expressed as:
\[ (x - 2)(x + 1) = x^2 - x - 2 = 0 \]Matching coefficients with \(x^2 + bx + c = 0\):
\[ b = -1, \quad c = -2 \]Correct answer: E
Question 9: Given \(X + Y = \dfrac{11}{5}\) and \(XY = \dfrac{2}{5}\), with \(X > 1\), find \(X\) and \(Y\).
Options:
- \(X = \dfrac{1}{5}, Y = 2\)
- \(X = 2, Y = \dfrac{1}{5}\)
- \(X = 2, Y = \dfrac{2}{5}\)
- \(X = \dfrac{3}{5}, Y = \dfrac{1}{5}\)
- \(X = 10, Y = \dfrac{1}{5}\)
Solution & Explanation:
From \(XY = \frac{2}{5}\), express \(Y\) in terms of \(X\):
\[ Y = \frac{2}{5X} \]Substitute into \(X + Y = \frac{11}{5}\):
\[ X + \frac{2}{5X} = \frac{11}{5} \]Multiply through by \(5X\):
\[ 5X^2 + 2 = 11X \quad \Rightarrow \quad 5X^2 - 11X + 2 = 0 \]Factor the quadratic:
\[ (5X - 1)(X - 2) = 0 \quad \Rightarrow \quad X = \frac{1}{5} \text{ or } X = 2 \]Since \(X > 1\), we choose \(X = 2\). Then find \(Y\):
\[ Y = \frac{11}{5} - 2 = \frac{1}{5} \]Correct answer: B
Question 10: Find \(x > 1\) such that \(x + \dfrac{1}{x} = \dfrac{10}{3}\)
Options:
- \(x = 6\)
- \(x = \dfrac{1}{3}\)
- \(x = 10\)
- \(x = 3\)
- \(x = 13\)
Solution & Explanation:
Multiply both sides by \(3x\) to eliminate fractions:
\[ 3x^2 + 3 = 10x \quad \Rightarrow \quad 3x^2 - 10x + 3 = 0 \]Factor the quadratic:
\[ (3x - 1)(x - 3) = 0 \quad \Rightarrow \quad x = \frac{1}{3} \text{ or } x = 3 \]Since \(x > 1\), the solution is \(x = 3\).
Correct answer: D