How to foil this equation...

jonnyboy17 · #1 $Old$ January 12th, 2009, 06:10 PM

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.

skeeter · #2 $Old$ January 12th, 2009, 06:22 PM

Mush · #3 $Old$ January 12th, 2009, 06:23 PM

Quote:

Originally Posted by jonnyboy17 $View Post$

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:

$(\sqrt{2x+2} - \sqrt{x-3})^2 = 4$

And we know that $(a+b)^2 = a^2+2ab+b^2$

Hence $(\sqrt{2x+2})^2 - 2\sqrt{x-3}\sqrt{2x+2}+(\sqrt{x-3})^2 = 4$

Now for the middle term use the rule $\sqrt{a}\sqrt{b} =\sqrt{ab}$ , and for the first and last terms use $\sqrt{a}^2 = a$ , which gives:

$2x+2 - 2\sqrt{(x-3)(2x+2)}+x-3 = 4$

Now you have to use foil on the term INSIDE the square root. Can you solve from here?