# Use Square Roots to Solve Quadratic Equations

## Key Questions

• If the question is about using the square root directly against the equation, the answer is definitely NO.

However, with certain transformation of a given equation into a different but equivalent form it is possible. Here is the idea.

Assume, for example, the same equation as analyzed in the previous answer:
${x}^{2} + x = 63$

If we could transform it to something like ${y}^{2} = b$ then the square root of both sides would deliver a solution.
So, let's transform our equation to this form.
Expression ${x}^{2} + x$ is not a square of anything, but ${x}^{2} + x + \frac{1}{4}$ is a square of $x + \frac{1}{2}$ because
${\left(x + \frac{1}{2}\right)}^{2} = {x}^{2} + 2 \cdot x \cdot \frac{1}{2} + \frac{1}{4} = {x}^{2} + x + \frac{1}{4}$

Therefore, it is reasonable to transform the original equation into
${\left(x + \frac{1}{2}\right)}^{2} - \frac{1}{4} = 63$ or
${\left(x + \frac{1}{2}\right)}^{2} = \frac{253}{4}$
From the last equation, which is absolutely equivalent to the original one, using the operation of the square root, we derive two linear equations:
$x + \frac{1}{2} = \frac{\sqrt{253}}{2}$ and $x + \frac{1}{2} = - \frac{\sqrt{253}}{2}$

So, two solutions are:
$x = \frac{- 1 + \sqrt{253}}{2}$ and $x = \frac{- 1 - \sqrt{253}}{2}$

The above method is pretty universal and handy if you don't remember a formula for solutions of a quadratic equation. Let me illustrate this with another example.
$- 3 {x}^{2} + 2 x + 8 = 0$

Step 1. Divide everything by $- 3$ to have ${x}^{2}$ with a multiplier $1$:
${x}^{2} - \frac{2}{3} x - \frac{8}{3} = 0$

Step 2. Since a coefficient at $x$ is $- \frac{2}{3}$, use ${\left(x - \frac{1}{3}\right)}^{2}$ in a transformed equation:
${\left(x - \frac{1}{3}\right)}^{2} - \frac{1}{9} - \frac{8}{3} = 0$ or
${\left(x - \frac{1}{3}\right)}^{2} = \frac{25}{9}$

Step 3. Use square root:
$x - \frac{1}{3} = \frac{5}{3}$ and $x - \frac{1}{3} = - \frac{5}{3}$

Step 4. Solutions:
$x = \frac{6}{3} = 2$ and $x = - \frac{4}{3}$

• Let us solve the following quadratic equation.

${x}^{2} - 6 x + 7 = 0$

by adding 2,

${x}^{2} - 6 x + 9 = 2$

${\left(x - 3\right)}^{2} = 2$

by taking the square-root,

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

by adding 3,

$x = 3 \pm \sqrt{2}$

I hope that this was helpful.