Quadratic Equations Practice Problems with Solutions
A set of multiple-choice practice questions on quadratic equations is presented below. Detailed, step-by-step explanations are included in the collapsible sections.
Practice Questions
Question 1
What are the two solutions to: \[ 2x^2 + 3x - 2 = 0 \]
A) \(-2 , 3\) B) \(-2 , -\frac{1}{2}\) C) \(2 , -\frac{1}{2}\) D) \(-2 , \frac{1}{2}\) E) \(-\frac{1}{2} , -2\)
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To factor \(2x^2 + 3x - 2\), find two numbers that multiply to \(a \cdot c = 2 \cdot (-2) = -4\) and add to \(b = 3\). These numbers are \(4\) and \(-1\).
Rewrite the middle term: \(2x^2 + 4x - 1x - 2 = 0\).
Factor by grouping: \(2x(x + 2) - 1(x + 2) = 0 \implies (2x - 1)(x + 2) = 0\).
Setting each factor to zero gives \(2x - 1 = 0 \implies x = 1/2\) and \(x + 2 = 0 \implies x = -2\).
Answer: D
Question 2
What is the sum of the two solutions to: \[ (x + 4)(x - 3) = 7 \]
A) \(-1\) B) \(-2\) C) \(1\) D) \(2\) E) \(3\)
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Expand the left side: \(x^2 - 3x + 4x - 12 = 7 \implies x^2 + x - 12 = 7\).
Set to standard form: \(x^2 + x - 19 = 0\).
Using Vieta's formulas, for \(ax^2 + bx + c = 0\), the sum of roots is \(-\frac{b}{a}\). Here \(b=1, a=1\), so sum = \(-1/1 = -1\).
Answer: A
Question 3
What is the product of the two solutions to: \[ (x - 2)(x - 6) = -3 \]
A) \(12\) B) \(-12\) C) \(15\) D) \(-3\) E) \(3\)
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Expand the left side: \(x^2 - 8x + 12 = -3\).
Set to standard form: \(x^2 - 8x + 15 = 0\).
Using Vieta's formulas, the product of roots is \(\frac{c}{a}\). Here \(c=15, a=1\), so product = \(15/1 = 15\).
Answer: C
Question 4
Find all values of \(m\) for which \( x^2 + 2x - 2m = 0 \) has no real solutions:
A) \(m = 0\) B) \(m = -2\) C) \(m < -4\) D) \(m = \frac{1}{2}\) E) \(m < -\frac{1}{2}\)
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A quadratic equation has no real solutions if the discriminant \(\Delta = b^2 - 4ac\) is less than 0.
Here \(a=1, b=2, c=-2m\). Therefore, \(\Delta = (2)^2 - 4(1)(-2m) = 4 + 8m\).
Setting \(\Delta < 0\): \(4 + 8m < 0 \implies 8m < -4 \implies m < -1/2\).
Answer: E
Question 5
Find \(m\) for which \( 2x^2 + 3x - m + 2 = 0 \) has two distinct real solutions:
A) \(m > 0\) B) \(m = 7\) C) \(m > \frac{7}{8}\) D) \(m < -\frac{7}{8}\) E) \(m = \frac{7}{8}\)
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For two distinct real roots, the discriminant \(\Delta = b^2 - 4ac\) must be greater than 0.
\(a=2, b=3, c=(-m+2)\).
\(\Delta = (3)^2 - 4(2)(-m+2) = 9 - 8(-m+2) = 9 + 8m - 16 = 8m - 7\).
Setting \(\Delta > 0\): \(8m - 7 > 0 \implies 8m > 7 \implies m > 7/8\).
Answer: C
Question 6
Which equation has two real solutions greater than zero?
A) \(x^2 + x = 0\) B) \(2x^2 - 10x = 28\) C) \(-x^2 + 4x + 5 = 0\) D) \(-3x^2 - 9 = -12x\) E) \(-3x^2 - 6x + 24 = 0\)
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For two positive real roots, sum (\(-b/a\)) > 0 AND product (\(c/a\)) > 0.
Checking (D) \(-3x^2 + 12x - 9 = 0\). Divide by -3: \(x^2 - 4x + 3 = 0\).
Sum = \(-(-4)/1 = 4\) (Positive). Product = \(3/1 = 3\) (Positive). Both roots are positive (1 and 3).
Answer: D
Question 7
Which has two real solutions whose product is greater than zero?
A) \(-x^2 - 2x = -8\) B) \(x^2 + 9x = -18\) C) \(-x^2 = -6 + x\) D) \(x^2 = 4x\) E) \(x^2 - 3x = 4\)
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Product of roots = \(c/a\). We need \(c/a > 0\).
Checking (B) \(x^2 + 9x + 18 = 0\). \(a=1, c=18\). Product = \(18/1 = 18 > 0\).
Answer: B
Question 8
\(b\) and \(c\) in \(x^2 + bx + c = 0\). Find \(b, c\) so solutions are \(x = -\frac{1}{4}, x = \frac{1}{2}\).
A) \(b = -\frac{1}{4}, c = -\frac{1}{8}\) B) \(b = -1, c = -1\) C) \(b = \frac{1}{4}, c = -1\) D) \(b = \frac{1}{4}, c = \frac{1}{8}\) E) \(b = 4, c = 8\)
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If roots are \(r_1, r_2\), the quadratic is \((x - r_1)(x - r_2) = 0\).
\((x - (-1/4))(x - 1/2) = (x + 1/4)(x - 1/2) = x^2 - 1/2x + 1/4x - 1/8 = x^2 - 1/4x - 1/8\).
Comparing to \(x^2 + bx + c\), we see \(b = -1/4\) and \(c = -1/8\).
Answer: A
Question 9
\(b\) and \(c\) in \(-x^2 + bx + c = 0\). Find \(b, c\) so sum is 6 and product is 8.
A) \(b = 6, c = 8\) B) \(b = -6, c = 8\) C) \(b = 8, c = -6\) D) \(b = 6, c = -8\) E) \(b = -8, c = 6\)
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Divide \(-x^2 + bx + c = 0\) by -1: \(x^2 - bx - c = 0\).
Sum of roots = \(-(-b)/1 = b\). We are given sum = 6, so \(b = 6\).
Product of roots = \(-c/1 = -c\). We are given product = 8, so \(-c = 8 \implies c = -8\).
Answer: D
Question 10
Which pair of quadratic equations are equivalent?
A) \(x^2 - 1 = 0\) & \(x^2 = -1\) B) \(-x^2 + x = -6\) & \(x^2 - 2x = 3\) C) ...
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Two equations are equivalent if one can be transformed into the other by multiplying or dividing by a non-zero constant.
Consider option (E): \(x^2 + x - 2 = 0\) and \(-2x^2 - 2x + 4 = 0\).
Multiply the first equation by -2: \(-2(x^2 + x - 2) = -2(0) \implies -2x^2 - 2x + 4 = 0\). They are identical equations.
Answer: E