sqrt(x-27) + sqrt(x) = 9
solve for x...
i tried squaring both sides to get the sqrts to go away but i still couldnt get the right answer.
I need to know how to get to the final solution. FYI the answer is 36.
Hello,
Square it :
$\displaystyle (x-27)+x+2 \sqrt{x(x-27)}=81$
$\displaystyle 2x+2 \sqrt{x(x-27)}=108$
$\displaystyle \sqrt{x(x-27)}=54-x$
Square again and solve the quadratic
Recall that $\displaystyle x-27 \ge 0$ and $\displaystyle x \ge 0$ (which makes $\displaystyle x \ge 27$)
Another way of doing it :
Let $\displaystyle a=\sqrt{x-27}$ and $\displaystyle b=\sqrt{x}$
We have $\displaystyle a+b=9$
Now, multiply both sides by $\displaystyle (a-b)$ :
$\displaystyle \underbrace{(a+b)(a-b)}_{a^2-b^2}=9(a-b)$
$\displaystyle (x-27)-x=9(a-b)$
$\displaystyle -27=9(a-b)$
$\displaystyle a-b=-3$
So you have :
$\displaystyle \left\{\begin{array}{ll} a+b=9 \quad (1) \\ a-b=-3 \quad (2) \end{array} \right.$
$\displaystyle (1)-(2) ~:~ 2b=12 \implies b=6 \implies \sqrt{x}=6 \implies x=36$
Check if it is correct :
$\displaystyle (1)+(2) ~:~ 2a=6 \implies a=3 \implies \sqrt{x-27}=3 \implies x-27=9 \implies x=36$
Yay !
Lookie here! You might like it this way, too. Pardon me, Moo. I just needed to post something. Haven't been on a lot today.
$\displaystyle \sqrt{x-27}+\sqrt{x}=9$ Move the $\displaystyle \sqrt{x}$ to the RHS.
$\displaystyle \sqrt{x-27}=9-\sqrt{x}$ Square both sides.
$\displaystyle x-27=81-18\sqrt{x}+x$ Collect terms.
$\displaystyle -108=-18\sqrt{x}$ Square both sides again.
$\displaystyle 11664=324x$ Divide
$\displaystyle \boxed{x=36}$