Higher Maths Question

[jodyemm]

Write f(x) = x² + 6x + 11 in the form (x+a)²+b

[Old_Geezer]

I'd reckon root six might be part of it.

[bibblebub]

(x+3)² + 2

[scotman]

(x + 3)^2 +2

[Old_Geezer]

Ah, well it's Monday and I'm not 100% awake, that's my excuse ;-)

What about jodyemm's other homework ?

[bibblebub]

must try harder - see me after class

[jodyemm]

Thank you :)

[bibblebub]

being given the answer is no replacement for knowing how to derive it

[jake-the-peg]

for that see completing the square here:

http://www.purplemath.com/modules/sqrquad.htm

[factor-fiction]

An alternative approach to completing the square is to use the identity:

x² + 6x + 11 Ξ (x+a)²+b and then try to find the values of a and b.

Expanding the right hand side of the identity tells us:
x² + 6x + 11 Ξ x² + a² +2ax +b
Comparing the coefficients of x tells us that 6 = 2a , so a=3
Comparing the constants tells us that 11= a² +b
And since we know a=3 then a² = 9 so b must equal 11-9=2

So our (x+a)²+b is (x+3)²+2

It's a long way round,, I know but it's useful if you forget how to complete the square