Cambridge 'Definite and Indefinite Quadratics', Discriminant (1 Viewer)

[Mumma]

Help...

Find the values of K for which
4a²-10ab+10b² + K(3a²-10ab+3b²)
is a perfect square.

I am not sure on how to proceed on this one. The farthest I could go was just factorising the inside of whats in the brackets.

The answers are -5/4 and 3/4

 

[acmilan]

4a² -10ab+10b² + K(3a² -10ab+3b² )
= (4+3k)a² - (10+10k)ab + (10+3k)b²

Recall how you expand a perfect square (ka - mb)² (where k and m are any numbers):

(ka - mb)² = k²a² - 2kmab + m²b²

the coefficient of ab = -2km = -2*root(coefficient of a²)*root(coefficient of b²)

So in the question, for it to be a perfect square:

coefficient of ab = -2*root(coeff. of a²)*root(coeff. of b²)
-(10+10k) = -2root(4+3k)root(10+3k)
(10+10k)² = 2(4+3k)(10+3k)
100 + 200k + 100k² = 4(4+3k)(10+3k)
100 + 200k + 100k² = 4(40 + 42k + 9k²)
100 + 200k + 100k² = 160 + 168k + 36k²
64k² + 32k - 60 = 0
16k² + 8k - 15 = 0

k = (-8 + 32)/32 or (-8-32)/32

k = -5/4 or 3/4

I hope that makes sense

edited to clarify it more for anyone else who doesnt understand it.

 

[Mumma]

Recall how you expand a perfect square (ka - mb)2 (where k and m are any numbers):

(ka - mb)2 = k2a2 - 2kmab + m2b2

Strange, I dont remember being taught that. Nothing in your working out has anything to do with the discriminant, so Im kinda wondering if that can be used in some way?

Cause I remember another question
2lx²+2lx+1 for values of l that make it a perfect square, I found the discriminate and made that equal to 0, then solved for l.

 

[Mumma]

Ah, i could probably do that for
(4+3k)a² - (10+10k)ab + (10+3k)b²
actually... Just needed to expand what was in the brackets after K.
Yeah turns out to be pretty much the same as your working out. Thanks for the help!