# Maths Question help

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#1
x^3 - 4x^2 - 3x +18=(x+a)(x-b)^2 for all x. Find the values of a and b.
I have expanded the brackets and compare co-effifient so i get the following
a-2b=-4
b^2 - 2ab=-3
ab^2 = 18
How can i find a and b from these equations?
0
6 months ago
#2
(Original post by George2111)
x^3 - 4x^2 - 3x +18=(x+a)(x-b)^2 for all x. Find the values of a and b.
I have expanded the brackets and compare co-effifient so i get the following
a-2b=-4
b^2 - 2ab=-3
ab^2 = 18
How can i find a and b from these equations?
What have you tried?
0
#3
My head is just frazzled and this question is really frustrating me, can you please give me some guidance?
0
6 months ago
#4
(Original post by George2111)
My head is just frazzled and this question is really frustrating me, can you please give me some guidance?
You could try rearranging the first to give a = ...

Then substitute into the second one ...
0
#5
Thanks for the response Muttley79
a=2b-4

b^2 - 2b(2b-4)=-3

-3b^2 +8b =-3

What do i do next?
0
6 months ago
#6
(Original post by George2111)
Thanks for the response Muttley79
a=2b-4

b^2 - 2b(2b-4)=-3

-3b^2 +8b =-3

What do i do next?
Sorry I had an online meeting!

I would collect it on the right hand side to give 3b^2 - 8b - 3 = 0 can you carry on from there?
0
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