# How do you solve x^2 - 21 = 3(5 - x^2)?

Mar 28, 2018

$x = \pm 3$

#### Explanation:

We want to expand the brackets firstly:

$3 \times 5 = 15$

$3 \times - {x}^{2} = - 3 {x}^{2}$

This, therefore:

${x}^{2} - 21 = 3 \left(5 - {x}^{2}\right) \to {x}^{2} - 21 = 15 - 3 {x}^{2}$

As we want to isolate the $x$ to one side, we do not want $- 3 {x}^{2}$, and therefore we do the opposite which is to $+ 3 {x}^{2}$. Notice that these cancel out. Also, remember what we do to one side we must do to another.

${x}^{2} - 21 = 15 - 3 {x}^{2} \to 4 {x}^{2} - 21 = 15$

We do not want $- 21$, as we would like to get $x$ on its own, and therefore do the opposite to $- 21$ which is to $+ 21$, notice the $- 21 + 21$ cancels out. Also, remember what you do to one side you MUST do to another.

$4 {x}^{2} - 21 = 15 \to 4 {x}^{2} = 36$

As we would like the value of $x$, first we realise we have $4$ lots of $x$ and divide both sides by $4$

$4 {x}^{2} = 36 \to {x}^{2} = 9$

As we would like $x$, we need to sqrt as that it the opposite to $^ 2$, notice the $^ 2$ cancels out. Also, remember to do this to both sides. We must also remember to take both square roots as they both satisfy the original equation

${x}^{2} = 9 \to x = \pm \sqrt{9} = \pm 3$

$\therefore$ $x = \pm 3$