Question

How do you solve by completing the square: $a{x}^2+bx+c=0$.

Answer 1

To solve the quadratic equation:

We have a quadratic equation in standard form $a{x}^2+bx+c=0$.

It is required to solve it for $x$ by completing the square.

$\begin{array}{rcl}a{x}^2+bx+c& =& 0\\ & \Rightarrow & a{x}^2+bx=-c\text{(Took the constant term to right side of equation)}\\ & \Rightarrow & x^2+\frac{b}{a}x=-\frac{c}{a}\text{(Divided both sides of equation by}a\text{)}\\ & \Rightarrow & x^2+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^2={\left(\frac{b}{2a}\right)}^2-\frac{c}{a}\text{(Added square of half the coefficient of}x\text{to both sides.)}\\ & \Rightarrow & {\left(x+\frac{b}{2a}\right)}^2=\frac{b^2}{4a^2}-\frac{c}{a}\text{(Applied the identity}{A}^2+2AB+B^2={\left(A+B\right)}^2\text{)}\\ & \Rightarrow & x+\frac{b}{2a}=\pm \sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}\\ & \Rightarrow & x=-\frac{b}{2a}\pm \sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}\\ & \Rightarrow & x=\frac{-b}{2a}\pm \sqrt{\frac{b^2-4ac}{4a^2}}\\ & \Rightarrow & x=\frac{-b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}\\ & \Rightarrow & x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$

Hence, the given quadratic equation $\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}\mathbf{=}\mathbf{0}$ is solved by completing the square.

The solutions are $\begin{array}{rcl}\mathit{x}& \mathbf{=}& \frac{\mathbf{-}\mathbf{b}\mathbf{+}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}\end{array}$ and $\begin{array}{rcl}\mathit{x}& \mathbf{=}& \frac{\mathbf{-}\mathbf{b}\mathbf{-}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}\end{array}$.