Algebra 1 Solving Quadratic Equations Worksheet Solutions (PDF)

Add 2 to both sides: \(3x^{2} – 5x + 2 = 0\).

\(a = 3, b = -5, c = 2\)

Take the square root of both sides: \(x = \pm\sqrt{49}\).

\(x = 7\) or \(x = -7\)

Add 5: \(x^{2} = 25\). Then \(x = \pm 5\).

\(x = 5\) or \(x = -5\)

Take square roots: \(x + 2 = \pm 5\).

\(x = -2 + 5 = 3\) or \(x = -2 – 5 = -7\).

\(x = 3\) or \(x = -7\)

Divide by 2: \(x^{2} = 9\).

\(x = 3\) or \(x = -3\)

\(x = \pm\sqrt{20} = \pm\sqrt{4 \cdot 5}\).

\(x = \pm 2\sqrt{5}\)

\((x + 2)(x + 3) = 0\).

Set each factor to 0: \(x = -2\) or \(x = -3\).

\(x = -2\) or \(x = -3\)

\((x – 3)(x – 4) = 0\).

\(x = 3\) or \(x = 4\)

Move everything to one side: \(x^{2} – 3x = 0\). Factor: \(x(x – 3) = 0\).

Dividing by \(x\) would lose the solution \(x = 0\).

\(x = 0\) or \(x = 3\)

Factor: \((2x – 1)(x + 3) = 0\).

\(2x – 1 = 0\) gives \(x = \frac{1}{2}\); \(x + 3 = 0\) gives \(x = -3\).

\(x = \frac{1}{2}\) or \(x = -3\)

Add \((6 \div 2)^{2} = 9\) to both sides: \(x^{2} + 6x + 9 = 16\).

\((x + 3)^{2} = 16\), so \(x + 3 = \pm 4\).

\(x = 1\) or \(x = -7\)

Move the constant: \(x^{2} – 4x = 5\). Add \((-4 \div 2)^{2} = 4\): \(x^{2} – 4x + 4 = 9\).

\((x – 2)^{2} = 9\), so \(x – 2 = \pm 3\).

\(x = 5\) or \(x = -1\)

\(a = 1, b = 4, c = 1\).

\(x = \frac{-4 \pm \sqrt{16 – 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2}\).

\(x = -2 \pm \sqrt{3}\)

\(a = 2, b = -3, c = -5\). Discriminant: \(9 + 40 = 49\).

\(x = \frac{3 \pm 7}{4}\), so \(x = \frac{10}{4} = \frac{5}{2}\) or \(x = \frac{-4}{4} = -1\).

\(x = \frac{5}{2}\) or \(x = -1\)

(a) \(36 – 36 = 0\): one solution. (b) \(1 – 24 = -23\): no real solutions. (c) \(9 + 16 = 25\): two solutions.

(a) 0, one; (b) -23, none; (c) 25, two

\(w(w + 3) = 40\), so \(w^{2} + 3w – 40 = 0\).

\((w + 8)(w – 5) = 0\), so \(w = -8\) or \(w = 5\). A width cannot be negative.

Width 5 feet, length 8 feet

Set \(-16t^{2} + 64t = 48\) and divide by \(-16\): \(t^{2} – 4t + 3 = 0\).

\((t – 1)(t – 3) = 0\), so \(t = 1\) or \(t = 3\).

The ball is at 48 feet on the way up and on the way down.

1 second and 3 seconds

\(n(n + 1) = 72\), so \(n^{2} + n – 72 = 0\).

\((n + 9)(n – 8) = 0\), so \(n = 8\) (positive).

8 and 9

\((x + 4)^{2} = 100\), so \(x + 4 = \pm 10\).

\(x = 6\) or \(x = -14\). A side cannot be negative.

6 feet

Set \(-2x^{2} + 50x – 100 = 200\), so \(-2x^{2} + 50x – 300 = 0\).

Divide by \(-2\): \(x^{2} – 25x + 150 = 0\), so \((x – 10)(x – 15) = 0\).

10 items or 15 items

\(b^{2} + (b + 7)^{2} = 13^{2}\), so \(2b^{2} + 14b + 49 = 169\).

\(2b^{2} + 14b – 120 = 0\), so \(b^{2} + 7b – 60 = 0\), which factors as \((b + 12)(b – 5) = 0\).

The base is \(b = 5\) feet and the height is \(5 + 7 = 12\) feet.

Base 5 feet, height 12 feet