Solution by Completing the Square

Solution by completing the square

Before taking up the work which follows, review the exercises in forming trinomial squares (the lesson on the square of a binomial).

Example 1. Solve x^2 + 6x - 16 = 0.

Check: substituting 2 for x: 4 + 12 - 16 = 0; substituting -8: 64 - 48 - 16 = 0.

The 9 added to each member of (2) is the square of half the coefficient of x; that is, 9 = \left(\frac62\right)^2, or 3^2. If the coefficient of x^2 is 1, the trinomial square can always be completed by adding the square of half the coefficient of x. If the coefficient of x^2 is -1 or any number other than +1, the equation is solved as in the next example.

The sign \pm properly belongs to each member of (5). Thus \pm(x + 3) = \pm 5. This, however, gives precisely the same roots as x + 3 = \pm 5. For this reason the sign \pm is put before one member only.

Graphed, the first member of (1) crosses the x-axis exactly at 2 and -8 (press Graph to see it):

y = x^2 + 6x - 16

Example 2. Solve 3x^2 - 7x - 20 = 0.

Transposing, 3x^2 - 7x = 20. Dividing by the coefficient of x^2, x^2 - \frac{7x}{3} = \frac{20}{3}. Adding \left(-\frac76\right)^2 to each member, x^2 - \frac{7x}{3} + \left(\frac76\right)^2 = \frac{20}{3} + \frac{49}{36} = \frac{289}{36}. Then \left(x - \frac76\right)^2 = \left(\frac{17}{6}\right)^2, so x - \frac76 = \pm\frac{17}{6}, and x = \frac76 \pm \frac{17}{6} = 4 or -\frac53.

Check: 3 \cdot 4^2 - 7 \cdot 4 - 20 = 48 - 28 - 20 = 0, and 3\left(-\frac53\right)^2 - 7\left(-\frac53\right) - 20 = \frac{25}{3} + \frac{35}{3} - 20 = 0.

Example 3. Solve 4x^2 - 4x - 79 = 0.

Transposing and dividing by 4, x^2 - x = \frac{79}{4}. Adding \left(-\frac12\right)^2, x^2 - x + \frac14 = \frac{80}{4}, \qquad \left(x - \tfrac12\right)^2 = 20. So x - \frac12 = \pm\sqrt{20} = \pm 2\sqrt5, and x = \frac12 \pm 2\sqrt5. Since \sqrt5 = 2.2360\ldots, the roots are 4.972\ldots and -3.972\ldots

Check: the decimal values are not the exact values of x; substituted for x, they make the first member nearly but not quite zero. An exact check substitutes the radical forms. Both at once: 4\left(\tfrac12 \pm 2\sqrt5\right)^2 - 4\left(\tfrac12 \pm 2\sqrt5\right) - 79 = 1 \pm 8\sqrt5 + 80 - 2 \mp 8\sqrt5 - 79 = 0. The radical terms vanish, because the two upper signs must be taken together, and then the two lower signs.

In quadratic equations like these, the radical forms of the roots are often sufficient; at other times values to two or three decimal places are necessary. Unless otherwise directed, obtain only the radical forms of irrational roots. Equations like 4x^2 - 4x - 79 = 0 can be solved graphically or by completing the square, but their solution by factoring is beyond us.

Rule

1. Transpose so that the terms containing x are in the first member and those which do not contain x are in the second.
2. Divide both members by the coefficient of x^2 (unless that coefficient is +1).
3. Add to both members the square of one half the coefficient of x (in the equation just obtained), making the first member a perfect trinomial square.
4. Rewrite the equation, expressing the first member as the square of a binomial and the second member in its simplest form.
5. Extract the square root of both members, writing the sign \pm before the square root of the second member. This gives two linear equations.
6. Solve the equation in which the second member has the sign +, and then the one with the sign -. The results are the roots of the quadratic.

Check: substitute each result separately for x in the original equation. If the resulting equations are not obvious identities, simplify each until it becomes one.

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: