# Thread: Algebraic Manipulations and Equations

1. ## Algebraic Manipulations and Equations

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.

2. Hello,
Originally Posted by johntuan
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.
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 $\displaystyle x \ge 27$)

3. 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 !

4. Originally Posted by Moo
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 $\displaystyle (a-b)$ :
$\displaystyle \underbrace{(a+b)(a-b)}_{a^2-b^2}=9(a-b)$

(x-27)-x=9(a-b)
-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 !
I'm probably being dumb once again, but where did those x's go?

5. Originally Posted by shinhidora
I'm probably being dumb once again, but where did those x's go?
$\displaystyle (x-27)-x=x-27-x=(x-x)-27=-27$

And I kept (a-b) in the left side because we don't need to transform them into x.

6. Originally Posted by Moo
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 $\displaystyle x \ge 27$)
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

$\displaystyle \boxed{x=36}$

7. Originally Posted by Moo
$\displaystyle (x-27)-x=x-27-x=(x-x)-27=-27$

And I kept (a-b) in the left side because we don't need to transform them into x.

I'm so stupid... I read the $\displaystyle (x-27)-x$ as $\displaystyle -x(x-27)$...

As I said, I was being dumb