Let p and q be real numbers such that p not equal0 , p^3 not equal q and p^3 not equal -q . if a and b are the non zero complex numbers satifying a + b = -p and a^3 + b^3 = q then a quadratic equation having roots a/b and b/a as its roots is?

Question

anonymous · Answer

With these roots equation form will be:

$$(bx-a)(ax-b) = 0$$

$$ab x^2 - (a^2 + b^2)x + ab = 0$$

anonymous · Answer

a and b are here complex numbers..

anonymous · Answer

The answer shiuld be      (p^3+ q)x^2  -  (p^3 - 2q) x + (p^3 + q) =0

anonymous · Answer

So we have to prove:

$$ab = p^3 + q$$

$$a^2 + b^2 = p^3 - 2q$$

anonymous · Answer

$$a^3+b^3=(a+b)((a+b)^2-3ab)$$

anonymous · Answer

$$ab = \frac{p^3 + q}{3p}$$

anonymous · Answer

$$a^2 +b^2 = (a+b)^2 - 2ab$$

anonymous · Answer

$$a^2+b^2=\frac{p^3-2q}{3p}$$

anonymous · Answer

Yep..

anonymous · Answer

we are done

anonymous · Answer

Yes we are done..

There was denominator too..

I could not guess as it became 0 after solving..

anonymous · Answer

Not 0 but 1..

anonymous · Answer

$p\neq0$  we can multiply by  $3p$

anonymous · Answer

Yes.. That is what I meant..

anonymous · Answer

It is given in question...

anonymous · Answer

@Yahoo! you have the values can you now substitute and solve ??

anonymous · Answer

@mukushla   @shubhamsrg    @waterineyes   ..  ab = p^2 - q /3

ajprincess · Answer

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Is it k nw?