Solving quadratic equations completing the square continued assignment

If you've got a quadratic equation on the form of

$$ax^{2}+c=0$$

Then you can solve the equation by using the square root of

$$x=\pm \sqrt{\frac{-c}{a}}$$

Example

$$3x^{2}-243=0$$

$$3x^{2}=243$$

$$x^{2}=\frac{243}{3}$$

$$x^{2}=81$$

$$x=\pm \sqrt{81}$$

$$x=9\: \: or\: \: x=-9$$

This method can only be used if b = 0. If we instead have an equation on the form of

$$x^{2}+bx=0$$

we can't use the square root initially since we do not have c-value. But we can add a constant d to both sides of the equation to get a new equivalent equation that is a perfect square trinomial. Remember that a perfect square trinomial can be written as

$$x^{2}+bx + d=\left ( x+d \right )^{2}=0$$

This process is called completing the square and the constant d we're adding is

$$d=\left (\frac{b}{2} \right )^{2}$$

Example

$$x^{2}+12x=0$$

We begin by finding the constant d that can be used to complete the square.

$$d=\left (\frac{b}{2} \right )^{2}=\left ( \frac{12}{2} \right )^{2}=6^{2}=36$$

$$x^{2}+12x+d=0+d\Rightarrow$$

$$x^{2}+12x+36=0+36\Rightarrow$$

$$\begin{pmatrix}x+6 \end{pmatrix}^{2}=36$$

$$\sqrt{\begin{pmatrix} x+6 \end{pmatrix}^{2}}=\pm \sqrt{36}$$

$$\begin{matrix} x+6=6\: \: &or\: \: & x+6=-6\\ x=0 & & x=-12 \end{matrix}$$

The completing the square method could of course be used to solve quadratic equations on the form of

$$ax^{2}+bx+c=0$$

In this case you will add a constant d that satisfy the formula

$$d=\left ( \frac{b}{2} \right )^{2}-c$$

Video lesson

Solve the equation by completing the squares

$$x^{2} - 3x - 10=0$$