• Dec 3rd 2005, 11:46 PM
coopsterdude
Hi Please help me to understand how squaring both sides of this equation gets the results on the second line. I have attached the problem. Thanks for any help.
• Dec 4th 2005, 12:52 AM
CaptainBlack
Let:

$\displaystyle \sqrt{x+8}\ =\ \sqrt{x+35}\ -\ 3$

squaring both sides gives:

$\displaystyle (\sqrt{x+8})^2\ =\ (\sqrt{x+35}\ -\ 3)^2$

or:

$\displaystyle x+8\ =\ (\sqrt{x+35}\ -\ 3)^2$

expanding the RHS:

$\displaystyle x+8\ =\ (\sqrt{x+35})^2\ +\ 2.(-3).\sqrt{x+35}\ +\ (-3)^2$

simplifying (a bit):

$\displaystyle x+8\ =\ x+35\ -\ 6.\sqrt{x+35}\ +\ 9$,

finaly:

$\displaystyle x+8\ =\ x\ +\ 49\ -\ 6.\sqrt{x+35}$,
• Dec 4th 2005, 01:05 AM
ticbol
Quote:

Originally Posted by coopsterdude
Hi Please help me to understand how squaring both sides of this equation gets the results on the second line. I have attached the problem. Thanks for any help.

I really don't know how to quote also the attached message.
The attached message says:

>>>
sqrt(x+8) = sqrt(x+35) -3
x+8 = x +35 -6sqrt(x+35) +9
<<<

Okay.
Let us forget the radicals first.
y = w -3
Square both sides,
y^2 = (w-3)^2
y^2 = w^2 -3w -3w +9
y^2 = w^2 -2(3w) +9
y^2 = w^2 -6w +9 -----------***

Now, if y were sqrt(x+8), and w = sqrt(x+35), what will happen?
This will happen:
[sqrt(x+8)]^2 = [sqrt(x+35)]^2 -6sqrt(x+35) +9
x+8 = x+35 -6sqrt(x+35) +9
And that is the same as what was in the attached message.

---------
Squaring a binomial:

by FOIL,

(x +y)^2
= (x+y)(x+y)
= x^2 +x*y +y*x +y^2
= x^2 +xy +xy +y^2
= x^2 +2xy +y^2

See that 2xy?
• Dec 4th 2005, 09:42 AM
coopsterdude
Tanks that totally solved my problem. You guys are great.