1. ## Algebra simplify

Hi just some algebra problems I got a bit confused about...

Expand and simplify (3a-b)(3b-a)

I got to: 9ab-3a-3b+ab?!

x²+5x
------
x²-25

And for this one I got to:

x²+5x
-----------
(x+5)(x-5)

Also:

x²+3x-10
-----------
x²+7x+10

And I got to:

x-2
----
x+2

And find values of p and q in the identity x²+6x+10=(x+p)² + p

Hence or otherwise, show that there are no real solutions of the equation x²+6x+10=0

Thanks

2. Hello,

Originally Posted by Turple
Hi just some algebra problems I got a bit confused about...

Expand and simplify (3a-b)(3b-a)

I got to: 9ab-3a-3b+ab?!
Yes... but 9ab+ab=... ?

x²+5x
------
x²-25

And for this one I got to:

x²+5x
-----------
(x+5)(x-5)
Hey, note that $x^2+5x=x(x+5)$

Also:

x²+3x-10
-----------
x²+7x+10

And I got to:

x-2
----
x+2
Ok

And find values of p and q in the identity x²+6x+10=(x+p)² + q

Hence or otherwise, show that there are no real solutions of the equation x²+6x+10=0
Hmm...

$x^2+6x+10=(x^2+6x+9)+1=(x+3)^2+1$

$x^2+6x+10=0 \Longleftrightarrow (x+3)^2+1=0 \Longleftrightarrow (x+3)^2=-1$

Is it possible ?

3. Originally Posted by Moo
Hmm...

$x^2+6x+10=(x^2+6x+9)+1=(x+3)^2+1$

$x^2+6x+10=0 \Longleftrightarrow (x+3)^2+1=0 \Longleftrightarrow (x+3)^2=-1$

Is it possible ?
I don't see why not!