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.

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- Dec 3rd 2005, 11:46 PMcoopsterdudePlease help with squaring booth sides
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 AMCaptainBlack
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 AMticbolQuote:

Originally Posted by**coopsterdude**

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.

Suppose we instead have:

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 AMcoopsterdude
Tanks that totally solved my problem. You guys are great.