How can I solve for z? 3+sqrt(z-6)=sqrt(z+9)

Printable View

- Jan 8th 2013, 05:09 PMEspionageBasic Square Root Help
How can I solve for z? 3+sqrt(z-6)=sqrt(z+9)

- Jan 8th 2013, 05:31 PMtopsquarkRe: Basic Square Root Help
This is going to be a problem in two steps. The base example here is how to solve the following for z:

$\displaystyle \sqrt{z + 1} + 2 = 3$

The general rule is to isolate the square root, then square both sides. In your problem we need to do this*twice*.

So

$\displaystyle 3 + \sqrt{z-6} = \sqrt{z+9}$

One square root is already isolated, so square both sides:

$\displaystyle 9 + 6 \sqrt{z - 6} + (z - 6) = z + 9$

Now isolate the square root:

$\displaystyle 6 \sqrt{z - 6} = 6$

And square both sides.

Always always always when you get a final answer for z make sure it is a solution to the original equation. Extra solutions tend to come out that don't satisfy the original equation.

-Dan - Jan 8th 2013, 05:41 PMEspionageRe: Basic Square Root Help
Where does the 6 come from when you square both sides the first time?

- Jan 8th 2013, 06:54 PMDevenoRe: Basic Square Root Help
$\displaystyle (3 + \sqrt{z - 6})^3 = 3^2 + (2)(3)(\sqrt{z - 6}) + (\sqrt{z - 6})^2 = 9 + 6\sqrt{z - 6} + (z - 6)$

because:

$\displaystyle (a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2$