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Cubic Roots Question (1 Viewer)

nrlwinner

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Hi. I have been stuck on this question for literally the past few days.

Find the cubic equation whose roots are twice those of the equation
 

tommykins

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3x^3/8 - 2x^2/4 + 1 = 0
3x^3 - 4x^2 + 8 = 0

Pretty sure there shouldn't be a +24
 

Tsylana

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Yeah tommy is right , no doubt. common 4u style question.
 

Aerath

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Can someone explain this question further for me.
It's really a 4U question. Pretty much, if the roots are a, b and c, the equation is 3x^3 - 2x^2 + 1 = 0.

That means that:
3a^3 - 2a^2 + 1 = 0, and so on and so forth.

Now, the roots were 2a, 2b, 2c:, the equation has to change, so that 2a, 2b and 2c are the new roots of the equation. To do this, we replace x by x/2.
3(x/2)^3 - 2(x/2)^2 + 1 = 0

This is so, when we sub in x = 2a:
3(2a/2)^3 - 2(2a/2)^2 + 1 = 0
and that equals
3a^3 - 2a^2 + 1 = 0

And we know that that equals zero.
 

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