Question

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

Answer 1

Explanation:

First, put the equation in standard quadratic form:

$(x-3{)}^2=5$

$(x-3)(x-2)=5$

$x^2-3x-3x+9=5$

$x^2-6x+9=5$

$x^2-6x+9-5=5-5$

$x^2-6x+4=0$

We can now use the quadratic equation to solve this problem:

The quadratic formula states:

For $a{x}^2+bx+c=0$, the values of $x$ which are the solutions to the equation are given by:

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

Substituting:

$1$ for $a$

$-6$ for $b$

$4$ for $c$ gives:

$x=\frac{--6\pm \sqrt{-6^2-\left(\mathrm{4.1.4}\right)}}{2.1}$

$x=\frac{6\pm \sqrt{36-16}}{2}$

$x=\frac{6\pm \sqrt{20}}{2}$

$x=\frac{6-\sqrt{4.5}}{2}$ ; $x=\frac{6+\sqrt{20}}{2}$

$x=\frac{6\pm \sqrt{4}\sqrt{5}2}{2}$ $x=\frac{6\pm \sqrt{4}\sqrt{5}}{2}$

$x=\frac{6\pm 2\sqrt{5}2}{2}$; $x=\frac{6\pm 2\sqrt{5}2}{2}$

$x=\frac{6}{2}-\frac{2\sqrt{5}}{2};\frac{6}{2}+\frac{2\sqrt{5}}{2}$

$x=3-\sqrt{5};x=3+\sqrt{5}$