# Can you use square roots to solve all quadratic equations?

Dec 22, 2014

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 $x = - \frac{4}{3}$