Quadratic Equation Practice Problems with Step-by-Step Solutions

Detailed solutions to quadratic equation problems are presented below.

1. Find the two solutions to the quadratic equation \(2x^2 + 3x - 2 = 0\).

   Solution

   Factor the left-hand side of the equation:

   \[ 2x^2 + 3x - 2 = (2x - 1)(x + 2) = 0 \]

   Solve each factor:

   \[ 2x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2} \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

   Solution set: \(\{-2, \frac{1}{2}\}\)

2. Find the sum of the two solutions to \((x + 4)(x - 3) = 7\).

   Solution

   The sum of the solutions of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(-\frac{b}{a}\).

   Rewrite the equation in standard form:

   \[ (x + 4)(x - 3) = 7 \] \[ x^2 + x - 12 = 7 \] \[ x^2 + x - 19 = 0 \]

   The sum of the solutions is:

   \[ -\frac{b}{a} = -\frac{1}{1} = -1 \]

3. Find the product of the two solutions to \((x - 2)(x - 6) = -3\).

   Solution

   The product of the solutions of \(ax^2 + bx + c = 0\) is given by \(\frac{c}{a}\).

   Rewrite in standard form:

   \[ (x - 2)(x - 6) = -3 \] \[ x^2 - 8x + 12 = -3 \] \[ x^2 - 8x + 15 = 0 \]

   The product of the solutions is:

   \[ \frac{c}{a} = \frac{15}{1} = 15 \]

4. Find all values of \(m\) for which \(x^2 + 2x - 2m = 0\) has no real solutions.

   Solution

   A quadratic equation has no real solutions when its discriminant is negative.

   The discriminant is:

   \[ \Delta = 2^2 - 4(1)(-2m) = 4 + 8m \]

   Set the discriminant less than zero:

   \[ 4 + 8m < 0 \] \[ 8m < -4 \] \[ m < -\frac{1}{2} \]

   The equation has no real solutions for \(m < -\frac{1}{2}\).

5. Find all values of \(m\) for which \(2x^2 + 3x - m + 2 = 0\) has two distinct real solutions.

   Solution

   Two distinct real solutions occur when the discriminant is positive.

   The discriminant is:

   \[ \Delta = 3^2 - 4(2)(-m + 2) = 9 + 8m - 16 = 8m - 7 \]

   Set the discriminant greater than zero:

   \[ 8m - 7 > 0 \] \[ 8m > 7 \] \[ m > \frac{7}{8} \]

   The equation has two distinct real solutions for \(m > \frac{7}{8}\).

6. Which quadratic 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) \(-3x^2 - 6x + 24 = 0\)

   Solution

   For both solutions to be positive, their sum and product must be positive.

   For each equation in standard form \(ax^2 + bx + c = 0\):

   • A) \(x^2 + x = 0\): Sum = \(-1\), Product = \(0\)
   • B) \(2x^2 - 10x - 28 = 0\): Sum = \(5\), Product = \(-14\)
   • C) \(-x^2 + 4x + 5 = 0\): Sum = \(4\), Product = \(-5\)
   • D) \(-3x^2 + 12x - 9 = 0\): Sum = \(4\), Product = \(3\)
   • E) \(-3x^2 - 6x + 24 = 0\): Sum = \(-2\), Product = \(-8\)

   Only equation D has both positive sum and positive product.

   Answer: D

7. Which quadratic equation has two real solutions with a product greater than zero?

   A) \(-x^2 - 2x = -8\)

   B) \(x^2 + 9x = -18\)

   C) \(-x^2 = -6 + x\)

   D) \(x^2 = 4x\)

   E) \(x^2 - 3x = 4\)

   Solution

   Find the product of solutions \(\frac{c}{a}\) for each:

   • A) \(-x^2 - 2x + 8 = 0\): Product = \(-8\)
   • B) \(x^2 + 9x + 18 = 0\): Product = \(18\)
   • C) \(-x^2 - x + 6 = 0\): Product = \(-6\)
   • D) \(x^2 - 4x = 0\): Product = \(0\)
   • E) \(x^2 - 3x - 4 = 0\): Product = \(-4\)

   Only equation B has a positive product.

   Answer: B

8. Find \(b\) and \(c\) in \(x^2 + bx + c = 0\) given solutions \(x = -\frac{1}{4}\) and \(x = \frac{1}{2}\).

   Solution

   A quadratic with given solutions can be written as:

   \[ \left(x + \frac{1}{4}\right)\left(x - \frac{1}{2}\right) = 0 \]

   Expand:

   \[ x^2 - \frac{1}{2}x + \frac{1}{4}x - \frac{1}{8} = 0 \] \[ x^2 - \frac{1}{4}x - \frac{1}{8} = 0 \]

   Thus, \(b = -\frac{1}{4}\) and \(c = -\frac{1}{8}\).

9. Find \(b\) and \(c\) in \(-x^2 + bx + c = 0\) given that the sum of solutions is 6 and their product is 8.

   Solution

   For \(-x^2 + bx + c = 0\) (or equivalently \(x^2 - bx - c = 0\)):

   Sum of solutions: \(\frac{-b}{-1} = b = 6\)

   Product of solutions: \(\frac{c}{-1} = -c = 8 \Rightarrow c = -8\)

   Thus, \(b = 6\) and \(c = -8\).

10. Which pair of quadratic equations are equivalent (have the same solutions)?

    A) \(x^2 - 1 = 0\) and \(x^2 = -1\)

    B) \(-x^2 + x = -6\) and \(x^2 - 2x = 3\)

    C) \(x^2 - 5x + 6 = 0\) and \(-x^2 - 5x - 6 = 0\)

    D) \(x^2 = 2x\) and \(x^2 + 2x = 0\)

    E) \(x^2 + x - 2 = 0\) and \(-2x^2 - 2x + 4 = 0\)

    Solution

    Rewrite each in standard form with positive leading coefficient:

    • A) \(x^2 - 1 = 0\) and \(x^2 + 1 = 0\) (different)
    • B) \(x^2 - x - 6 = 0\) and \(x^2 - 2x - 3 = 0\) (different)
    • C) \(x^2 - 5x + 6 = 0\) and \(x^2 + 5x + 6 = 0\) (different)
    • D) \(x^2 - 2x = 0\) and \(x^2 + 2x = 0\) (different)
    • E) \(x^2 + x - 2 = 0\) and \(x^2 + x - 2 = 0\) (same, after dividing the second by \(-2\))

    Answer: E

More References and Links

Quadratic Equations Tutorial
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