Hi All
This is my question.
I don't know how to begin working on it.
I already tried simplifying the first part to: \sqrt{63} = \sqrt{7 \cdot 3^3}=\sqrt{7}\sqrt{3^2}=3\sqrt{7}
But this doesn't get me closer to answering the first part of the question, and I think the same technique will...
Hi All
I have this problem. My workings are attached.
My answer was:
x=4
y=5
z=6
This is wrong. I don't know where I'm going wrong.
I'd be really grateful if someone could help.
Thanks!
Daniel
Hi All
I have the following question.
I have reviewed my notes but have not been able to crack this.
I tried two different ways, both wrong.
First:
(4-6*\sqrt2)^2=
16-24*\sqrt2-24*\sqrt2+(36*2)
= 88-218*\sqrt2
so, $x=88$ and
$y=218$My second method was
(4-6*\sqrt2)^2= 4^2-(6*2)=28...
Hi all
I'm trying to work out how to answer this type of problem.
(6x+2z)^2-64=(ax+2z+8)(-8+bx+cz) where a, b and c > 0
I have attempted the problem by expanding the brackets:
=36x^2+24xz+4z^2-64
This is the same as (6x+2z)^2-(8)^2
Then subtracting from either 'side' of the quadratic and...