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.
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:
$\displaystyle x+8\ =\ x\ +\ 49\ -\ 6.\sqrt{x+35}$,
I really don't know how to quote also the attached message.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.
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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?