How do you solve 4sqrtx=sqrt( 9x+9)?

May 2, 2018

$x = \frac{9}{7}$

Explanation:

First, square both sides:
${\left(4 \sqrt{x}\right)}^{\textcolor{b l u e}{2}} = {\left(\sqrt{9 x + 9}\right)}^{\textcolor{b l u e}{2}}$

$16 x = 9 x + 9$

Subtract $\textcolor{b l u e}{9} x$ from both sides of the equation:
$16 x \quad \textcolor{b l u e}{- \quad 9 x} = 9 x + 9 \quad \textcolor{b l u e}{- \quad 9 x}$

$7 x = 9$

Finally, divide both sides by $7$:
$\frac{7 x}{\textcolor{b l u e}{7}} = \frac{9}{\textcolor{b l u e}{7}}$

$x = \frac{9}{7}$

However, we should check our work by plugging our solution back into the original equation:
$4 \sqrt{\frac{9}{7}} = \sqrt{9 \left(\frac{9}{7}\right) + 9}$

$4 \cdot \frac{3}{\sqrt{7}} = \sqrt{\frac{81}{7} + \frac{63}{7}}$

$\frac{12}{\sqrt{7}} = \sqrt{\frac{144}{7}}$

$\frac{12}{\sqrt{7}} = \frac{12}{\sqrt{17}}$

Both sides of the equation equal, meaning that $x = \frac{9}{7}$ is true.

Hope this helps!

$x = \frac{9}{7}$.

Explanation:

First of All,

$\textcolor{w h i t e}{\times x} 4 \sqrt{x} = \sqrt{9 x + 9}$

$\Rightarrow 16 x = 9 x + 9$ [Square both sides]

$\Rightarrow 16 x - 9 x = 9$ [Transpose $9 x$ to L.H.S.]

$\Rightarrow 7 x = 9$

$\Rightarrow x = \frac{9}{7}$.

So, $x = \frac{9}{7}$.

But if we substitute $x = \frac{9}{7}$ in the equation, we get,

L.H.S. = $4 \sqrt{\frac{9}{7}} = \sqrt{16 \cdot \frac{9}{7}} = \sqrt{\frac{144}{7}}$

R.H.S = $\sqrt{9 \cdot \frac{9}{7} + 9} = \sqrt{\frac{81}{7} + 9} = \sqrt{\frac{81 + 63}{7}} = \sqrt{\frac{144}{7}}$

So, L.H.S. = R.H.S.

So $x = \frac{9}{7}$ is indeed a solution for this equation.

Hope this helps.