equation w. radicals and exponenets

$\displaystyle \sqrt{2x+2}-\sqrt{x-3}=2$

$\displaystyle \left(\sqrt{2x+2}\right)^{2}=\left(2+\sqrt{x-3}\right)^{2}$

$\displaystyle x+1=4\sqrt{x-3}$

i dont know where to go from here, i tried dividing by 4 and mult. the left by itself, and get lost, the answer is 7

Re: equation w. radicals and exponenets

Quote:

Originally Posted by

**mathmathmathmathmathmathm** $\displaystyle \sqrt{2x+2}-\sqrt{x-3}=2$

$\displaystyle \left(\sqrt{2x+2}\right)^{2}=\left(2+\sqrt{x-3}\right)^{2}$

$\displaystyle x+1=4\sqrt{x-3}$

i dont know where to go from here, i tried dividing by 4 and mult. the left by itself, and get lost, the answer is 7

$\displaystyle \displaystyle \begin{align*} \sqrt{2x+2} - \sqrt{x-3} &= 2 \\ \sqrt{2x+2} &= 2+\sqrt{x-3} \\ \left(\sqrt{2x+2}\right)^2 &= \left(2+\sqrt{x-3}\right)^2 \\ 2x+2 &= 4 + 4\sqrt{x-3} + x - 3 \\ x + 1 &= 4\sqrt{x-3} \\ (x + 1)^2 &= \left(4\sqrt{x-3}\right)^2 \\ x^2 + 2x + 1 &= 16(x - 3) \\ x^2 + 2x + 1 &= 16x - 48 \\ x^2 - 14x + 49 &= 0 \\ (x - 7)^2 &= 0 \\ x - 7 &= 0 \\ x &= 7 \end{align*}$

Re: equation w. radicals and exponenets

You should of course, check the solution. (Squaring both sides of an equation may introduce "extraneous" solutions.)

$\displaystyle \sqrt{2(7)+ 2}= \sqrt{14+ 2}= \sqrt{16}= 4$

$\displaystyle \sqrt{7- 3}= \sqrt{4}= 2$

4- 2= 2 so the solution checks.

Re: equation w. radicals and exponenets

thanks so much prove it, but can you explain when you get the (x-7) squared, how do you elimate the second (x-7) in the last step? does it cancel itself out? dont understand that step, thanks again

Re: equation w. radicals and exponenets

Quote:

Originally Posted by

**mathmathmathmathmathmathm** thanks so much prove it, but can you explain when you get the (x-7) squared, how do you elimate the second (x-7) in the last step? does it cancel itself out? dont understand that step, thanks again

To undo a square, you need to square root both sides. The square root of 0 is 0.