Question

How do you solve ${\left(x-3\right)}^2=5$?

Answer 1

Step- 1: Simplify the equation:

Given equation is ${\left(x-3\right)}^2=5$.

Use ${\left(a-b\right)}^2=a^2-2ab+b^2$

$\begin{array}{rcl}{\left(\text{x}-3\right)}^2& =& 5\\ {\text{x}}^2-6x+9& =& 5\\ {\text{x}}^2-6x+4& =& 0\end{array}$

Step- 2: Apply quadratic formula:

Value of $\text{x}$, for equation $a{x}^2+bx+c=0$ is given by quadratic formula as follows,

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

So, $a=1,b=-6,c=4$

$\begin{array}{rcl}\text{x}& =& \frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}\\ & =& \frac{6\pm \sqrt{36-16}}{2}\\ & =& \frac{6\pm \sqrt{20}}{2}\\ & =& \frac{6\pm 2\sqrt{5}}{2}\\ & =& 3\pm \sqrt{5}\end{array}$

Hence,value of $\mathbf{\text{x}}$ is $3\pm \sqrt{5}$.