4
[ v(2x+2) - v(x-3)][ v(2x+2) - v(x-3)] = 4
*By the way , that v symbol means SQuare root. CAn someplease walk me through this process.
Both terms in brackets are identical, so we can write this as:
$\displaystyle (\sqrt{2x+2} - \sqrt{x-3})^2 = 4 $
And we know that $\displaystyle (a+b)^2 = a^2+2ab+b^2$
Hence $\displaystyle (\sqrt{2x+2})^2 - 2\sqrt{x-3}\sqrt{2x+2}+(\sqrt{x-3})^2 = 4 $
Now for the middle term use the rule $\displaystyle \sqrt{a}\sqrt{b} =\sqrt{ab} $, and for the first and last terms use $\displaystyle \sqrt{a}^2 = a $, which gives:
$\displaystyle 2x+2 - 2\sqrt{(x-3)(2x+2)}+x-3 = 4 $
Now you have to use foil on the term INSIDE the square root. Can you solve from here?