Algebraic Manipulations and Equations

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September 11th 2008, 10:21 AM
johntuan

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.
September 11th 2008, 10:29 AM
Moo

Hello,

Quote:

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 :

$(x-27)+x+2 \sqrt{x(x-27)}=81$

$2x+2 \sqrt{x(x-27)}=108$

$\sqrt{x(x-27)}=54-x$

Square again and solve the quadratic :)
Recall that $x-27 \ge 0$ and $x \ge 0$ (which makes $x \ge 27$ )
September 11th 2008, 11:04 AM
Moo

Another way of doing it :

Let $a=\sqrt{x-27}$ and $b=\sqrt{x}$

We have $a+b=9$

Now, multiply both sides by $(a-b)$ :
$\underbrace{(a+b)(a-b)}_{a^2-b^2}=9(a-b)$

$(x-27)-x=9(a-b)$
$-27=9(a-b)$
$a-b=-3$

So you have :
$\left\{\begin{array}{ll} a+b=9 \quad (1) \\ a-b=-3 \quad (2) \end{array} \right.$

$(1)-(2) ~:~ 2b=12 \implies b=6 \implies \sqrt{x}=6 \implies x=36$

Check if it is correct :
$(1)+(2) ~:~ 2a=6 \implies a=3 \implies \sqrt{x-27}=3 \implies x-27=9 \implies x=36$

Yay !
September 11th 2008, 11:42 AM
shinhidora

Quote:

Originally Posted by Moo

Another way of doing it :

Let $a=\sqrt{x-27}$ and $b=\sqrt{x}$

We have $a+b=9$

Now, multiply both sides by $(a-b)$ :
$\underbrace{(a+b)(a-b)}_{a^2-b^2}=9(a-b)$

(x-27)-x=9(a-b)
-27=9(a-b)
$a-b=-3$

So you have :
$\left\{\begin{array}{ll} a+b=9 \quad (1) \\ a-b=-3 \quad (2) \end{array} \right.$

$(1)-(2) ~:~ 2b=12 \implies b=6 \implies \sqrt{x}=6 \implies x=36$

Check if it is correct :
$(1)+(2) ~:~ 2a=6 \implies a=3 \implies \sqrt{x-27}=3 \implies x-27=9 \implies x=36$

Yay !

I'm probably being dumb once again, but where did those x's go? :p
September 11th 2008, 11:48 AM
Moo

Quote:

Originally Posted by shinhidora

I'm probably being dumb once again, but where did those x's go? :p

$(x-27)-x=x-27-x=(x-x)-27=-27$

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

(Rofl)
September 11th 2008, 11:50 AM
masters

Quote:

Originally Posted by Moo

Hello,

Square it :

$(x-27)+x+2 \sqrt{x(x-27)}=81$

$2x+2 \sqrt{x(x-27)}=108$

$\sqrt{x(x-27)}=54-x$

Square again and solve the quadratic :)
Recall that $x-27 \ge 0$ and $x \ge 0$ (which makes $x \ge 27$ )

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.

$\sqrt{x-27}+\sqrt{x}=9$ Move the $\sqrt{x}$ to the RHS.

$\sqrt{x-27}=9-\sqrt{x}$ Square both sides.

$x-27=81-18\sqrt{x}+x$ Collect terms.

$-108=-18\sqrt{x}$ Square both sides again.

$11664=324x$ Divide

$\boxed{x=36}$
September 11th 2008, 11:53 AM
shinhidora

Quote:

Originally Posted by Moo

$(x-27)-x=x-27-x=(x-x)-27=-27$

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

(Rofl)

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

As I said, I was being dumb :p