# How do you evaluate 3 + sqrt(x+7) - sqrt(3x) when x = 9?

May 10, 2018

$\implies 7 - 3 \sqrt{3}$

#### Explanation:

$3 + \sqrt{x + 7} - \sqrt{3 x}$

When $x = 9$, we get

$3 + \sqrt{9 + 7} - \sqrt{3 \cdot 9}$

$= 3 + \sqrt{16} - \sqrt{3 \cdot {3}^{2}}$

$= 3 + 4 - 3 \sqrt{3}$

$= 7 - 3 \sqrt{3}$

May 10, 2018

$7 - 3 \cdot {\left(3\right)}^{\frac{1}{2}}$

#### Explanation:

Put $9$ in place of $x$

$3 + {\left(9 + 7\right)}^{\frac{1}{2}} - {\left(3 \cdot 9\right)}^{\frac{1}{2}}$

It will be

$3 + {16}^{\frac{1}{2}} - {27}^{\frac{1}{2}}$

As

${16}^{\frac{1}{2}} = 4$

Thus

$3 + 4 - {27}^{\frac{1}{2}}$

It will be

$7 - {27}^{\frac{1}{2}}$

As ${27}^{\frac{1}{2}}$ can be simplified to ${\left(3 \cdot 9\right)}^{\frac{1}{2}}$ and as $3 \cdot 9 = 27$, it will be the same as

${3}^{\frac{1}{2}} \cdot {9}^{\frac{1}{2}}$

Since

${9}^{\frac{1}{2}} = 3$

It will be

${3}^{\frac{1}{2}} \cdot 3$

Coming back to

$7 - {27}^{\frac{1}{2}}$

As

${27}^{\frac{1}{2}} = 3 \cdot {3}^{\frac{1}{2}}$

So it will be

$7 - 3 \cdot {3}^{\frac{1}{2}}$

May 10, 2018

$\implies 7 - 3 \sqrt{3}$

#### Explanation:

Plugging in $9$ for $x$, we get

$3 + \sqrt{9 + 7} - \sqrt{3 \cdot 9}$

which simplifies to

$3 + \sqrt{16} - \sqrt{27}$

$\sqrt{16}$ evaluates to $4$. Thus we have

$3 + 4 - \sqrt{27}$

$\implies 7 - \sqrt{27}$

We can rewrite $\sqrt{27}$ as $\sqrt{9} \cdot \sqrt{3}$. We get

$7 - \left(\sqrt{9} \cdot \sqrt{3}\right)$

$\implies 7 - 3 \sqrt{3}$

Hope this helps!