GMAT Quadratic Equation Problems
with Solutions and Explanations – Sample 3
Detailed step-by-step solutions to the quadratic equation problems in Sample 3, similar to GMAT test questions.
Solution to Question 1
Given the equation:
\[
x^2 + 5x + 6 = 0
\]
Factor the quadratic:
\[
(x + 2)(x + 3) = 0
\]
Solve each factor:
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
Solution: x = -2, -3
Solution to Question 2
Given solutions x = 1 and x = -2 satisfy:
\[
2x^2 + kx - m = 0
\]
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:
\[
(k - m) - (-2k - m) = -2 - (-8) \quad \Rightarrow \quad 3k = 6 \quad \Rightarrow \quad k = 2
\]
\[
k - m = -2 \quad \Rightarrow \quad 2 - m = -2 \quad \Rightarrow \quad m = 4
\]
Solution: k = 2, m = 4
Solution to Question 3
\[
(x - 1)(x + 3) = 1 - x
\]
Bring all terms to one side:
\[
(x - 1)(x + 3) - (1 - x) = 0 \quad \Rightarrow \quad (x - 1)(x + 3) + (x - 1) = 0
\]
Factor:
\[
(x - 1)((x + 3) + 1) = (x - 1)(x + 4) = 0
\]
Solve:
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]
\[
x + 4 = 0 \quad \Rightarrow \quad x = -4
\]
Solution: x = 1, -4
Solution to Question 4
\[
2 - (x - 2)^2 = -18
\]
Subtract 2:
\[
-(x - 2)^2 = -20
\]
Multiply by -1:
\[
(x - 2)^2 = 20
\]
Take the square root:
\[
x - 2 = \pm \sqrt{20} = \pm 2\sqrt{5} \quad \Rightarrow \quad x = 2 \pm 2\sqrt{5}
\]
Solution to Question 5
For a single solution, discriminant must be zero:
\[
x^2 + kx + 4 = 0 \quad \Rightarrow \quad k^2 - 16 = 0 \quad \Rightarrow \quad k = 4
\]
Solution to Question 6
\[
(m + 2)x^2 = 16, \quad x = 2
\]
\[
4(m + 2) = 16 \quad \Rightarrow \quad m + 2 = 4 \quad \Rightarrow \quad m = 2
\]
\[
4x^2 = 16 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = \pm 2
\]
Second solution: x = -2
Solution to Question 7
\[
\frac{1}{2}a x^2 + 3x - 4 = 0, \quad x = -2
\]
\[
\frac{1}{2}a(-2)^2 + 3(-2) - 4 = 0 \quad \Rightarrow \quad 2a - 10 = 0 \quad \Rightarrow \quad a = 5
\]
\[
\frac{5}{2}x^2 + 3x - 4 = 0 \quad \Rightarrow \quad (x + 2)\left(\frac{5}{2}x - 2\right) = 0
\]
\[
\frac{5}{2}x - 2 = 0 \quad \Rightarrow \quad x = \frac{4}{5}
\]
Solution to Question 8
\[
(x - 2)(x + 1) = x^2 - x - 2
\]
\[
b = -1, \quad c = -2
\]
Solution to Question 9
We are given: \[ X \cdot Y = \frac{2}{5} \]
Express \( Y \) in terms of \( X \): \[ \quad Y = \frac{2}{5X} \]
Substitute into \( X + Y = \frac{11}{5} \) to obtain
\[
X + \frac{2}{5X} = \frac{11}{5}
\]
Multiply through by \( 5X \) to eliminate the fraction:
\[
5X^2 + 2 = 11X
\]
Bring all terms to one side: \[ 5X^2 - 11X + 2 = 0 \]
Factor the quadratic: \[ (5X - 1)(X - 2) = 0 \]
\[
5X - 1 = 0 \quad \Rightarrow \quad X = \frac{1}{5}
\]
\[
X - 2 = 0 \quad \Rightarrow \quad X = 2
\]
\( X \) is greater than 1, hence \( X = 2 \)
Now find the corresponding \( Y \) value using \( X + Y = \frac{11}{5} \):
If \( X = 2 \), then \[ Y = \frac{11}{5} - 2 = \frac{11}{5} - \frac{10}{5} = \frac{1}{5} \]
Thus, the two numbers are \[ X = 2 \text{ and } Y = \frac{1}{5}\]
Solution to Question 10
The reciprocal of \(x\) is \(\frac{1}{x}\). The sum of \(x\) and \(\frac{1}{x}\) is equal to \(\frac{10}{3}\), which can be written mathematically as:
\[
x + \frac{1}{x} = \frac{10}{3}
\]
Multiply both sides by \(3x\) to eliminate the fraction:
\[
3x^2 + 3 = 10x
\]
Bring all terms to one side to set the equation to 0:
\[
3x^2 - 10x + 3 = 0
\]
Factor the quadratic:
\[
(3x - 1)(x - 3) = 0
\]
The two solutions are:
\[
x = \frac{1}{3} \quad \text{and} \quad x = 3
\]
Since we are looking for the solution greater than 1, the final solution is:
\[
x = 3
\]
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