Polynomials (1 Viewer)

Smilebuffalo · Mar 14, 2010

If ax^3 + cx + d = 0 has a double root, show that 27ad^2 + 4c^3 = 0

bleaver · Mar 14, 2010

$ax^3+cx+d=0\\ \therefore 3ax^2+c=0\ (Will\ share\ a\ root)\\ x^2=\frac{-c}{3a}\\\\ x=\sqrt{\frac{-c}{3a}}$

We can then substitute this root into the original equation.
$a(\sqrt{\frac{-c}{3a}})^3+c(\sqrt{\frac{-c}{3a}})=-d\\ Factorising.. \sqrt{\frac{-c}{3a}}(\frac{-c}{3}+c)=-d\\ Squaring\ both\ sides...\\ \frac{-c}{3a}(\frac{2c}{3})^2=d^2\\ \frac{-4c^3}{27a}=d^2\\ 27ad^2+4c^3=0\\ As\ required.$

I've skipped a few lines of simplification but work thruogh it and it should be obvious.

Smilebuffalo · Mar 14, 2010

how come when you rooted it, you ignored the negative root?

bleaver · Mar 14, 2010

Think about it. If I put the negative root as a possibility, once I square both sides, that negative is gone leaving you with the same result as the positive possibility.

Polynomials (1 Viewer)

Smilebuffalo

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bleaver

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Smilebuffalo

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bleaver

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