Solution by Formula

Solution by formula

The equation ax^2 + bx + c = 0 is the general quadratic equation in standard form. Solving it by completing the square, exactly as in the last lesson, gives x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \qquad \text{(F)}

The value (F) is a general result and may be used as a formula to solve any quadratic equation. The solution of a quadratic by formula requires less labor than any other method, except for such equations as can be solved by factoring at sight. Those with considerable experience in algebra seldom solve a quadratic by any other method.

Example 1. Solve 3x^2 - 5x = 8.

Writing in standard form, 3x^2 - 5x - 8 = 0. Then 3 corresponds to a, -5 to b, and -8 to c. Substituting in (F), x = \frac{-(-5) \pm \sqrt{25 - 4 \cdot 3(-8)}}{2 \cdot 3} = \frac{5 \pm \sqrt{25 + 96}}{6} = \frac{5 \pm 11}{6} = \frac83 \text{ or } -1.

Example 2. Solve 2k^2x^2 = kx + 1.

Writing in standard form, 2k^2x^2 - kx - 1 = 0, so a = 2k^2, b = -k, and c = -1: x = \frac{-(-k) \pm \sqrt{(-k)^2 - 4 \cdot 2k^2(-1)}}{2 \cdot 2k^2} = \frac{k \pm \sqrt{9k^2}}{4k^2} = \frac{k \pm 3k}{4k^2} = \frac1k \text{ or } -\frac{1}{2k}.

Check: 2k^2\left(\frac1k\right)^2 = k\left(\frac1k\right) + 1, or 2 = 1 + 1; and 2k^2\left(-\frac{1}{2k}\right)^2 = k\left(-\frac{1}{2k}\right) + 1, or \frac12 = -\frac12 + 1.

When b^2 - 4ac is negative, the square root in (F) is imaginary: the graph of y = ax^2 + bx + c then does not meet the x-axis (type none for "no real root").

Typing roots

Separate the roots with commas: 2, -8. A pair of roots with a double sign may be typed with ± or +-: 1 ± sqrt(5) means 1 + \sqrt5 and 1 - \sqrt5, and (3 +- sqrt(17))/4 works too. Give irrational roots in simplest radical form: \sqrt{20} is marked "equal, but not simplified"; write 2\sqrt5. Where decimal values are asked for, type them to three places.

Try one: