The method of completing the square can be used to develop a formula that enables us to solve any quadratic equation. Consider the general quadratic equation :

$a^{2}+bx+c=0\text{ and }a\neq 0$

By applying the method of completing the square, we can obtain the solutions in terms of the coefficients a, b, and c.

$ax^2+bx+c=0$

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

$x^2+\frac{b}{a}x=-\frac{c}{a}$

Since the coefficient of x is b/a, the square of one – half the coefficient of x is $(\frac{b}{2a})^2$ to each side.

$x^{2}+\frac{b}{a}x+\left ( \frac{b}{2a} \right )^{2}=-\frac{c}{a}+\left ( \frac{b}{2a} \right )^{2}$

$\left ( x+\frac{b}{2a} \right )^{2}=-\frac{4ac}{4a^{2}}+\frac{b^{2}}{4a^{2}}$

$\left ( x+\frac{b}{2a} \right )^{2}=\frac{b^{2}-4ac}{4a^{2}}$

$x+\frac{b}{2a}=\pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}$

$x+\frac{b}{2a}=\pm \frac{\sqrt{b^{2}-4ac}}{2a}$

$x=-\frac{b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$

$\text{x}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

We now have a formula for solving any quadratic equation. It is usually stated as follows :

The solutions of the quadratic equation

$a^{2}+bx+c=0\text{ and }a\neq 0$

Are given by

$\text{x}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Example 1 :

Solve

$3x^2-4x=-1$

Using the quadratic formula.

Solution :

We first put the equation in standard form in order to identify a, b, and c.

$3x^2-4x=-1$

$3x^2-4x+1=0$

a = 3, b = – 4, c = 1.

Substituting these values for a, b, and c into the quadratic formula, we have :

$\text{x}=\frac{-(-4)\pm\sqrt{(-4)^{2}-4(3)(1)}}{2(3)}$

$\text{x}=\frac{4\pm\sqrt{16-12}}{6}$

$\text{x}=\frac{4\pm\sqrt{4}}{6}$

$\text{x}=\frac{4\pm2}{6}$

Therefore,

$\text{x}=\frac{4+2}{6}=1\text{ or }\text{x}=\frac{4-2}{6}=\frac{1}{3}$

The solution are 1 and 1/3.

Example 2 # :

Solve

$x^2+x-1=0$

Solution :

$x^2+x-1=0$

a = 1, b = 1, c = -1

$\text{x}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$\text{x}=\frac{-1\pm\sqrt{(1)^{2}-4(1)(-1)}}{2(1)}$

$\text{x}=\frac{-1\pm\sqrt{1+4}}{2}$

$\text{x}=\frac{-1\pm\sqrt{5}}{2}$

The solution are

$\frac{-1+\sqrt{5}}{2}\text{ and }\frac{-1-\sqrt{5}}{2}$

Example 3 # :

Solve

$\frac{x^2}{6}+\frac{x}{3}-\frac{1}{3}=0$

Solution :

We first simplify the equation by multiplying both sides by the lcd 6.

$\frac{x^2}{6}+\frac{x}{3}-\frac{1}{3}=0$

$6\left ( \frac{x^2}{6}+\frac{x}{3}-\frac{1}{3} \right )=6.0$

$x^2+2x-2=0$

a = 1, b = 2, c = – 2

$\text{x}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$\text{x}=\frac{-2\pm\sqrt{(2)^{2}-4(1)(-2)}}{2(1)}$

$\text{x}=\frac{-2\pm\sqrt{12}}{2}$

$\text{x}=\frac{-2\pm2\sqrt{3}}{2}$

Dividing the denominator 2 into both terms in the numerator simplifies the solution to

$\text{x}=\frac{-2}{2}\pm\frac{2\sqrt{3}}{2}$

$\text{x}=-1\pm\sqrt{3}$

The solutions are

$-1+\sqrt{3}\text{ and }-1-\sqrt{3}$

Example 4# :

Solve

$x^2-7x=0$

Solution :

$x^2-7x=0$

a = 1, b = -7 , c = 0

$\text{x}=\frac{-(-7)\pm\sqrt{(2)^{2}-4(1)(-2)}}{2(-2)}$

$\text{x}=\frac{7\pm\sqrt{49-0}}{2}$

$\text{x}=\frac{7\pm7}{2}$

$\text{x}=\frac{7+7}{2}\text{ or }\text{x}=\frac{7-7}{2}$

x = 7 or x = 0

The solutions are 7 and 0.

Example 5# :

Solve

$x^2+2x+5=0$

Solution :

$x^2+2x+5=0$

a = 1, b = 2, c = 5

$\text{x}=\frac{-2\pm\sqrt{(2)^2-4(1)(5)}}{2(1)}$

$\text{x}=\frac{-2\pm\sqrt{4-20}}{2}$

$\text{x}=\frac{-2\pm\sqrt{-16}}{2}$

But $\sqrt{-16}$ is not a real number. Therefore this quadratic equation has no real number solution.

The method of solving a quadratic equation by completing the square can be rather tedious, and for that reason it is seldom used to solve a quadratic equation. The operation of completing the square does, however, have a variety of other application in algebra.

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