# How do you solve 7x^2-6=57?

Oct 3, 2017

$x = + 3 \mathmr{and} x = - 3$

#### Explanation:

There is no term in $x$, so you can solve for ${x}^{2}$

$7 {x}^{2} = 57 + 6 \text{ } \leftarrow$ add 6 to both sides.

$7 {x}^{2} = 63 \text{ } \leftarrow \div 7$

${x}^{2} = 9$

$x = \pm \sqrt{9}$

$x = \pm 3$

Oct 3, 2017

Solution: $x = 3 , x = - 3$

#### Explanation:

$7 {x}^{2} - 6 = 57 \mathmr{and} 7 {x}^{2} = 57 + 6 \mathmr{and} 7 {x}^{2} = 63$ or

${x}^{2} = \frac{63}{7} \mathmr{and} {x}^{2} = 9 \mathmr{and} x = \pm \sqrt{9} \mathmr{and} x = \pm 3$

Solution: $x = 3 , x = - 3$ [Ans]

Oct 3, 2017

$x = 3$ or $x = - 3$

#### Explanation:

First, we add $6$ to both sides.
$7 {x}^{2} - 6 = 57$
$7 {x}^{2} - 6 \textcolor{b l u e}{+} \textcolor{b l u e}{6} = 57 \textcolor{b l u e}{+} \textcolor{b l u e}{6}$
Simplify:
$7 {x}^{2} = 63$

Divide both sides by $7$

$\frac{7 {x}^{2}}{7} = \frac{63}{7}$

Therefore, ${x}^{2} = 9$

If ${x}^{2} = a$, then $x = \sqrt{a}$ or $- \sqrt{a}$.

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

or

$x = - \sqrt{9}$
$x = - \sqrt{{3}^{2}}$
$x = - 3$

Both ways work.

Oct 3, 2017

$x = \pm 3$

#### Explanation:

If
$\textcolor{w h i t e}{\text{XXX}} 7 {x}^{2} - 6 = 57$
then
$\textcolor{w h i t e}{\text{XXX}} 7 {x}^{2} = 63$ ...after adding $6$ to both sides
and
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = 9$ ...after dividing both sides by $7$

Therefore
color(white)("XXX")x=+3" or " x=-3 ...after taking the square root of both sides (since ${\left(\pm 3\right)}^{2} = 9$