Math · College algebra · Worked example

Solve a quadratic with the quadratic formula

Solve 2x² + 3x − 1 = 0. Give the exact solutions and their values to three decimal places.

2 x^{2} + 3 x - 1 = 0

Check the form and read a, b and c

The equation is already in standard form, with 0 on the right. Read each coefficient with its sign: a = 2, b = 3 and c = −1. No pair of integers factors 2x² + 3x − 1, so the formula is the right tool.

a = 2, b = 3, c = - 1

Compute the discriminant

Find b² − 4ac first; it tells you what kind of answer to expect. The discriminant is 17, which is positive and not a perfect square, so there are two different irrational solutions.

b^{2} - 4 a c = 3^{2} - 4 (2) (- 1) = 9 + 8 = 17

\begin{aligned} b^{2} - 4 a c & = 3^{2} - 4 (2) (- 1) \\ = 9 + 8 = 17 \end{aligned}

Substitute into the formula

Put −b, the discriminant and 2a into the formula. The ± records both solutions at once, and the whole numerator stays over 2a = 4.

x = \frac{- 3 \pm \sqrt{17}}{4}

Separate the two solutions

The + sign and the − sign give the two solutions. Since √17 ≈ 4.1231, they are about 0.281 and −1.781 to three decimal places.

x = \frac{- 3 + \sqrt{17}}{4} \approx 0.281, x = \frac{- 3 - \sqrt{17}}{4} \approx - 1.781

\begin{array}{c} x = \frac{- 3 + \sqrt{17}}{4} \approx 0.281 \\ x = \frac{- 3 - \sqrt{17}}{4} \approx - 1.781 \end{array}

Compare with the graph

The parabola y = 2x² + 3x − 1 opens upward and crosses the x-axis twice, near x = 0.281 and x = −1.781, as the positive discriminant predicts. Its vertex lies halfway between them, at x = −3/4.

Result

x = (−3 + √17)/4 ≈ 0.281 or x = (−3 − √17)/4 ≈ −1.781.

x = \frac{- 3 \pm \sqrt{17}}{4}

Your turn

Solve x² − 4x − 1 = 0 with the quadratic formula and simplify the radical.

Show the answer and explanation

x = 2 ± √5.

Here a = 1, b = −4 and c = −1, so b² − 4ac = 16 + 4 = 20 and x = (4 ± √20)/2. Since √20 = 2√5, this is (4 ± 2√5)/2 = 2 ± √5, about 4.236 and −0.236.

\begin{array}{l} x^{2} - 4 x - 1 = 0 \\ x = \frac{4 \pm \sqrt{20}}{2} \\ x = 2 \pm \sqrt{5} \end{array}

Keep exploring

Open the graph and change the constant term from −1 to +2: the discriminant becomes 9 − 16 = −7 and the parabola lifts off the x-axis.

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