# If one root of the equation $ax^2+bx+c=0$ be the square of the other.

If $a \neq 0$ and if one root of the equation $ax^2+bx+c=0$ is the square of the other, prove that: $$b^3+a^2c+ac^2=3abc.$$

My Attempt: Given: $$ax^2 + bx + c=0$$ Let $\alpha$ and $\beta$ be the roots of the equation. $$\alpha + \beta = \dfrac {-b}{a}$$ $$\alpha . \beta = \dfrac {c}{a}$$ According to the question: $$\alpha = \beta^2$$

• What if $a=0\,$? If your question relies on additional assumptions, please spell them out. – dxiv Aug 19 '17 at 5:46
• @dxiv, There are no more assumptions. The question is exactly the same what I have typed. – pi-π Aug 19 '17 at 6:02
• If $a=0, b=1, c=-1$ then the only root is $x=1$ which is obviously its own square. However $b^2+a^2c+ac^2 = 1 \ne 0 = 3abc\,$, so the conclusion you propose doesn't follow. – dxiv Aug 19 '17 at 6:05
• @dxiv, What is the meaning of quadratic equation if $a=0$? – pi-π Aug 19 '17 at 6:07
• Where was quadratic mentioned before this last comment? All I am getting at is that's important that you make sure to give the full context to a question in order to elicit unambiguous answers (same as here). – dxiv Aug 19 '17 at 6:08

So, we have $$\dfrac ca=\alpha\beta=\beta^3$$

$$-\dfrac ba=\alpha+\beta=\beta^2+\beta$$

Cube both sides using $$(p+q)^3=p^3+q^3+3pq(p+q)$$

Replace the values of $\beta^3,\beta^2+\beta$

• Cubing both sides of which equation? – pi-π Aug 19 '17 at 6:03
• @blue_eyed_..., Not equation, but the relation : $$-\dfrac ba=\beta^2+\beta$$ – lab bhattacharjee Aug 19 '17 at 6:04

$$\alpha + \beta = \dfrac {-b}{a}$$ $$\alpha . \beta = \dfrac {c}{a}$$ Substitute $$\alpha = \beta^2$$ into previous equations,

$$a(\beta^2+\beta) = a\beta(\beta+1) = - b\tag{1}$$

$$a\beta^3 = c\tag{2}$$

Cube equation $(1)$,

$$a^3 \beta^3 (\beta^3 + 3 \beta^2 + 3 \beta + 1) = -b^3$$ Using equation $(2)$, $$a^2 c (\beta^3 + 3 \beta^2 + 3 \beta + 1) = -b^3$$

$$a c (a\beta^3 + 3a\beta( \beta + 1) + a) = -b^3$$

Using equation $(1)$ and $(2)$,

$$a c (c - 3b + a) = -b^3$$

$$ac^2-3abc+a^2c=-b^3$$

$$b^3+ac^2+a^2c = 3abc$$

Alternatively: $$\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)^2=\frac{-b+\sqrt{b^2-4ac}}{2a} \Rightarrow$$ $$b^2-2ac+ab=(a-b)\sqrt{b^2-4ac} \Rightarrow$$ Squaring and then dividing both sides by $4a$ will prove the claim.