Hey guys,

how can i expand this???

(k^2+1)x^2+2(5k-3)x+9=0

Thanks

Dan

ps. x^2 means x to the power of 2

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- Dec 28th 2012, 01:15 AMDanthemathsHard Expanding Brackets
Hey guys,

how can i expand this???

(k^2+1)x^2+2(5k-3)x+9=0

Thanks

Dan

ps. x^2 means x to the power of 2 - Dec 28th 2012, 02:17 AMSMADRe: Hard Expanding Brackets
well when you expand it you are supposed to get k^2x^2+x^2+10xk-6x+9=0

- Dec 28th 2012, 03:23 AMastartleddeerRe: Hard Expanding Brackets
Expansion is the opposite of factorisation.

Consider,

$\displaystyle x + x^2 + x^3 \Rightarrow x(1 + x + x^2) $

Have you even made an attempt to do a google search so as to see worked examples?

By the way, I'll make the assumption you want to solve for 'k' in this problem. Thus you will need to use the discriminant. I'll set it up for you, but the rest is down to you.

$\displaystyle [(2)(5k -3)]^2 - [(4)(k^2 + 1)(9)] > 0 $

I don't know what the quesion is, but there will be a high-probability the discriminant will be greater than zero for two real roots. If the equation has one real root only, then convert the "greater than" sign to an "equals."