# How do you solve \sqrt{x}=x-6?

May 2, 2018

$x = 9$

#### Explanation:

$\sqrt{x} = x - 6$
Square the equation:
$x = {\left(x - 6\right)}^{2}$
Apply the expansion of ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$
$\implies x = {x}^{2} - 12 x + 36$
$\implies 0 = {x}^{2} - 13 x + 36$
$\implies {x}^{2} - 9 x - 4 x + 36 = 0$
$\implies x \left(x - 9\right) - 4 \left(x - 9\right) = 0$
$\implies \left(x - 4\right) \left(x - 9\right) = 0$
$\implies x = 4 \mathmr{and} x = 9$

Note that substituting 4 in the equation returns 2 = -2, which is obviously wrong. So we neglect x = 4 in the set of solutions. Take care to verify your answers after solving(don't make my mistake!)

May 2, 2018

$x = 9$

#### Explanation:

$\sqrt{x} = x - 6$

First, square both sides:
${\sqrt{x}}^{\textcolor{red}{2}} = {\left(x - 6\right)}^{\textcolor{red}{2}}$

Simplify:
$x = {x}^{2} - 12 x + 36$

Move everything to one side of the equation:
$0 = {x}^{2} - 13 x + 36$

Now we need to factor.
Our equation is standard form, or $a {x}^{2} + b x + c$.

The factored form is $\left(x - m\right) \left(x - n\right)$, where $m$ and $n$ are integers.

We have two rules to find $m$ and $n$:

• $m$ and $n$ have to multiply up to $a \cdot c$, or $36$
• $m$ and $n$ have to add up to $b$, or $- 13$

Those two numbers are $- 4$ and $- 9$. So we put them into our factored form:

$0 = \left(x - 4\right) \left(x - 9\right)$

Therefore,

$x - 4 = 0$ and $x - 9 = 0$

$x = 4$ $\quad \quad \quad$ and $\quad \quad \quad$  x = 9

$- - - - - - - - - - - - - - - - - - - -$

However, we still need to check our answers by substituting them back into the original equation, since we have a square root in our original equation.

Let's first check if $x = 4$ is really a solution:
$\sqrt{4} = 4 - 6$

$2 = - 2$

This is not true! That means that $x \ne 4$ ($4$ is not a solution)

Now let's check $x = 9$:
$\sqrt{9} = 9 - 6$

$3 = 3$

This is true! That means that $x = 9$ ($9$ is really a solution)

So the final answer is $x = 9$.

Hope this helps!

May 2, 2018

$x = 9$ is the only real solution to this equation.

#### Explanation:

First, square both sides of this equation.

$x = {x}^{2} - 12 x + 36$

Now put in standard form.

${x}^{2} - 13 x + 36 = 0$

Factor.

$\left(x - 4\right) \left(x - 9\right) = 0$

$x = 9$ is a solution to this equation. $x = 4$ is not a solution to the original equation. However it is a solution to

$x = {x}^{2} - 12 x + 36$

When we squared both sides to at the beginning, we enabled an extraneous solution since ${\left(- \sqrt{x}\right)}^{2} = {\left(\sqrt{x}\right)}^{2} = x$. Thus we enabled $- \sqrt{x}$ as a valid left-hand side of the equation when the original problem did not. Note that $- \sqrt{x} = x - 6$ when $x = 4$, but this is not what the problem is asking.