(3x+4)^(1/2)-(2x+4)^(1/2)=2

sGt HarDKorE · Sep 22, 2010, 07:42 PM

My friend is helping his friend with math when they came to this question. They called me asking for help and I stumped. I subtracted 2 to the other side and then tried to expand. I am at a loss on what to do next.

sGt HarDKorE · Sep 22, 2010, 07:50 PM

Hi, please also read the rules. I am asking for help on what the next step is if I'm even going in the right direction. Please point out where I said anything about someone doing it for me? Please familiarize yourself with the rules! Thanks :)

Unknown008 · Sep 23, 2010, 02:58 AM

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

Well... remove the square roots, they're the major stumbling block here. Square both sides:

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

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

Clean it out.

$2\sqrt{6x^2+20x + 16}= 4+5x$

Hm... square again? :D

$(2\sqrt{6x^2+20x+16})^2= (4+5x)^2$

$4(6x^2+20x+16)= 16 + 40x + 25x^2$

$24x^2 + 80x + 64 = 16 + 40x + 25x^2$

$x^2 - 40x - 48 = 0$

Now, use the quadratic formula:

$x = \frac{40\pm \sqrt{40^2 - 4(1)(-48)}}{2(1)}$

$x = \frac{40 \pm \sqrt{1792}}{2}$

$x = \frac{40}{2} \pm \frac{16\sqrt{7}}{2}$

$x = 20 \pm 8\sqrt{7}$

Now, if you tried reinserting the negative value of x in the original equation, you'll get about -0.5, so this solution is rejected and the only solution is:

$x = 20 + 8\sqrt{7}$