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So, I have this question which is still troubling me:

Find the value of $k$ such that the equation $2x^3 + 3x^2 + kx - 48 = 0$ has two solutions equal in value but opposite in sign.

I've had numerous attempts at this, such as using simultaneous equations and the factor theorem, but there always seems to be a problem. I'm sure I'm missing an important step here. Any clearing up would be great, thanks!

Karl Kroningfeld
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missiledragon
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3 Answers3

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HINT:

So, we can assume the roots to be of the form $a,-a,b$

So using Vieta's formula

$a+(-a)+b=-\frac32\implies b=-\frac32$ and $a\cdot(-a)\cdot b=\frac{48}2\implies a^2=16\implies a=\pm4$

So, the roots are $\pm4,-\frac32$

Again using Vieta's formula $k=a\cdot b+(-a)\cdot b+a\cdot(-a)$

lab bhattacharjee
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Let $a$ be one of the pair roots. Then $$2a^3 + 3a^2 + ka - 48 =0=2(-a)^3 + 3(-a)^2 + k(-a) - 48\Rightarrow(2a^2+k)a=0\Rightarrow k=-2a^2$$ Substitute back this to the equation one gets $$2a^3+3a^2+(-2a^2)a-48=0$$which gives $a^2=16\Rightarrow k=-2a^2=-32$.

Easy
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well you can use comparison.

$2x^3+3x^2+kx−48=(2x+a)(x+b)(x-b) = (2x+a)(x^2-b^2) = (2x^3 +ax^2 - 2b^2x - ab^2)$

so a comparison

$2x^3+3x^2+kx−48 = (2x^3 +ax^2 - 2b^2x - ab^2)$

gives us a = 3, b = 4, k=-32

so answers the question would be $(2x+3)(x-4)(x+4)$

sorry i am not sure of how to make superscript for power.

asun
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