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If one root of the equation $ax^2+bx+c=0$ is the square of the other prove that $b^3+ac(c+a)=3abc$ I couldn't understand how to start the problem I considered the two roots as $p$ and $p^2$.

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Hint: If you considered the two roots as $p$ and $p^2$, then you can write the quadratic as $$ax^2 + bx + c = a(x-p)(x-p^2)$$


Alternatively, for a quadratic, using the formula for the sum and products of roots we have $$p + p^2 = -\frac{b}{a}$$ and $$p \cdot p^2 = p^3 = \frac{c}{a}$$ Now you just need to find a way to eliminate $p$ from the two equations.


Another way to do it would be to solve the quadratic as $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Since one root is the square of the other, then $$\frac{-b-\sqrt{b^2-4ac}}{2a} = \left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)^2$$

After re-arranging, squaring, simplifying and then dividing throughout by $a \neq 0$ we get $$b^3 + a^2c+ac^2 = 3abc$$

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  • $\begingroup$ How are you going to prove $\endgroup$ – Drake Benz Jul 10 '15 at 12:22
  • $\begingroup$ I don't know how to carry it further $\endgroup$ – Drake Benz Jul 10 '15 at 12:23
  • $\begingroup$ Zain it would be of great help if you'd how me how $\endgroup$ – Drake Benz Jul 10 '15 at 12:24

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