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If one root of the equation $ax^2+bx+c=0$ be the square of the other then which is true?

$1$. $a^3+b^3+c^3-3abc=0$

$2$. $a^3+b^3+bc^2=3abc$

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

$4$. none.

My Attempt:

Let one root be $\alpha $ then the other root will be $\alpha^2$. Then, $$(x-\alpha)(x-\alpha^2)=0$$ $$x^2-x(\alpha^2+\alpha)+\alpha^3=0$$ Comparing with $ax^2+bx+c=0$ we get, $$a=1$$ $$b=-(\alpha^2+\alpha)$$ $$c=\alpha^3$$

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  • $\begingroup$ You can't conclude a = 1. After all if $27 (3-x)=0$ and $42 (3-x)=0$ we can't conclude $27=42$. $\endgroup$ – fleablood Apr 26 '17 at 8:08
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Vieta's formula says:

$\alpha+\alpha^2=-\dfrac ba,\alpha\cdot\alpha^2=\dfrac ca$

$$\left(-\dfrac ba\right)^3=(\alpha+\alpha^2)^3=\alpha^3+(\alpha^3)^2+3\alpha^3\left(-\dfrac ba\right)$$

Replace $\alpha^3$ with $\dfrac ca$ and simplify

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Let one root be $\alpha $ then the other root will be $\alpha^2$. Then, $$(x-\alpha)(x-\alpha^2)=0$$

Note that you need a factorization of the form $\color{red}{a}(x-\alpha)(x-\alpha^2)=0$ to be able to compare with $ax^2+bx+c=0$. Alternatively, you can divide by $a$ and continue with a quadratic where $a=1$.

Other than that, the approach is fine. Insert the expressions you find for $a$, $b$ and $c$ into the three options. But carefully looking at the powers, you can already see 1. (because of $c^3$, leaving a term in $\alpha^9$) and 2. (because of $b^2c$, leaving a term in $\alpha^7$) won't simplify to $0$. That leaves calculating expression 3 and concluding it's either 3. or 4.


Alternatively, still calling the roots $\alpha$ and $\alpha^2$, we know that the sum and product of the roots of $ax^2+bx+c=0$ are equal to $-\tfrac{b}{a}$ and $\tfrac{c}{a}$ respectively, so: $$-\frac{b}{a}=\alpha+\alpha^2 \quad \mbox{and} \quad \frac{c}{a}=\alpha^3$$ And continue from there.

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  • $\begingroup$ Conclusion $a=1$ is wrong. $\endgroup$ – Wojowu Apr 26 '17 at 7:52
  • $\begingroup$ @Wojowu You mean OP's conclusion? Good catch, I'll edit accordingly. $\endgroup$ – StackTD Apr 26 '17 at 7:57
  • $\begingroup$ Yes, that's what I meant. Undownvoted :) $\endgroup$ – Wojowu Apr 26 '17 at 8:00
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Please, in your future exercises, try not to name different things with the same variable ($a$ in this example), because it causes a huge confusion. Also, remember than the quadratic factorization has the coefficient of $x^2$.

We have the equation : $ax^2 + bx + c = 0$

Now, let's assume it has two roots, $ρ_1 , ρ_2 $ with $ρ_2 = ρ_1^2$

Then, the polynomial will be factorized as :

$a(x-ρ_1)(x-ρ_2) = 0 \Leftrightarrow a(x-ρ_1)(x-ρ_1^2)=0 \Leftrightarrow ax^2 - aρ_1 (ρ_1 + 1) x + aρ_1 ^3 = 0 $

You can continue on from there.

Alternativelly, you can take on Vietta's Formulas :

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

$$\alpha\cdot\alpha^2=\dfrac ca$$

and continue on from there.

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  • $\begingroup$ Although I admit it is visually unclear, OP used $a$ in the equation and $\alpha$ for one of the roots, which are two distinct notations. $\endgroup$ – jvdhooft Apr 26 '17 at 8:05
  • $\begingroup$ Probably he did, which is true two distinct notations, but it is visually unclear nonetheless and causes so much confusion. It's always nice to have a clear presentation without things that confuse the reader. $\endgroup$ – Rebellos Apr 26 '17 at 8:10
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Product of roots $= \alpha^3 = \dfrac{c}{a}$

Now $a\alpha^2+b\alpha+c = 0 \Rightarrow a^3 \alpha^6+b^3 \alpha^3+c^3 = 3abc \alpha^3$

Substitute for $\alpha^3$ in the above and after simplification we obtain $b^3+a^2c+ac^2=3abc$

In the above we assume that $\alpha \ne 0$, but its easy to see that the above relation holds even when $\alpha = 0$ and hence all cases are covered.

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  • $\begingroup$ Your answer is correct but how do you get the implication in the second line? $\endgroup$ – Georges Elencwajg Apr 26 '17 at 20:20
  • $\begingroup$ I used the identity $a+b+c=0 \Rightarrow a^3+b^3+c^3=3abc$ $\endgroup$ – Hari Shankar Apr 27 '17 at 4:00

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