# How do you solve x= sqrt(6x) +7?

Apr 15, 2018

$x = 10 + \sqrt{51}$

#### Explanation:

$x = \sqrt{6 x} + 7$
First let's define the domain :
$x \ge 0$
So:
$x - 7 = \sqrt{6 x}$
(x-7)²=6x
(x²-14x+49)-6x=0
x²-20x+49=0
Δ=20²-4*49
Δ=204
x=(-b+-sqrtΔ)/2a
${x}_{\text{1}} = \frac{20 - \sqrt{204}}{2}$
${x}_{\text{2}} = \frac{20 + \sqrt{204}}{2}$
${x}_{\text{1}} = \frac{20 - 2 \sqrt{51}}{2}$
${x}_{\text{2}} = \frac{20 + 2 \sqrt{51}}{2}$
${x}_{\text{1}} = 10 - \sqrt{51} < 0 \implies$ NOT a solution.
${x}_{\text{2}} = 10 + \sqrt{51}$
So : $x = 10 + \sqrt{51}$

Apr 15, 2018

$x = 10 + \sqrt{51}$

Bare with me please: this is a long solution process:

#### Explanation:

We can subtract $7$ from both sides to arrive at:

$x - 7 = \sqrt{6 x}$

We can square both sides to get:

$\textcolor{b l u e}{{\left(x - 7\right)}^{2}} = 6 x$

$= \textcolor{b l u e}{\left(x - 7\right) \left(x - 7\right)} = 6 x$

What I have in blue, we can use the highly useful mnemonic FOIL to simplify this. We simply multiply the first terms, outside terms, inside and last terms. We get:

• First terms: $x \cdot x = {x}^{2}$
• Outside terms: $x \cdot - 7 = - 7 x$
• Inside terms: $- 7 \cdot x = - 7 x$
• Last terms: $- 7 \cdot - 7 = 49$

Now, we have:

$\textcolor{b l u e}{{x}^{2} - 7 x - 7 x + 49} = 6 x$

Which can be simplified to

${x}^{2} - 14 x + 49 = 6 x$

We have a quadratic on the left, so we want to set it equal to zero to find its zeroes. We do this by subtracting $6 x$ from both sides to get:

${x}^{2} - 20 x + 49 = 0$

The only factors of $49$ are $1 , 7$ and $49$, and neither of them, positive or negative, add up to $- 20$. We can use the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

where a=1, b=-20 & c=49 (in the form $a {x}^{2} + b x + c = 0$)

Plugging into the quadratic formula, we get:

$x = \frac{20 \pm \sqrt{400 - 4 \left(1 \cdot 49\right)}}{2 \cdot 1}$

Which simplifies to

$x = \frac{20 \pm \textcolor{b l u e}{\sqrt{204}}}{2}$

Because $51 \cdot 4 = 204$, we can rewrite $\textcolor{b l u e}{\sqrt{204}}$ as

$\textcolor{b l u e}{\sqrt{4} \cdot \sqrt{51}}$

Which obviously simplifies to $\textcolor{b l u e}{2 \sqrt{51}}$

Now, we have

$x = \frac{20 \pm 2 \sqrt{51}}{2}$

Every term is divisible by $2$, so we can factor a 2 out to get

$x = \frac{\textcolor{red}{10} \cancel{20} \pm \cancel{2} \sqrt{51}}{\textcolor{red}{1} \cancel{2}}$

Which is equal to

$\textcolor{red}{x = 10 \pm \sqrt{51}}$

Since our answer had a square root in it, we know the domain has to be $x \ge 0$.

$10 + \sqrt{51} > 0$, so this is a solution for $x$ but

$10 - \sqrt{51} < 0$, which is outside of the domain, so we get

$x = 10 + \sqrt{51}$

as our final solution.

This was a long solution process, but all I did was:

• Get $x$ on one side
• Square both sides to get rid of the square root
• Used FOIL to simplify the left side