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If the roots of given Quadratic equation $$a(b-c)x^2 +b(c-a)x + c(a-b)=0$$ are equal, prove the following: $$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$$.

MY approach:

Method 1: put Discriminant=0 and get stuck.

Method2:add $acx$ on both sides and get $(x-1)$ as factor ,so other root is also 1 and hence the result.
But if someone can prove the result by making discriminant zero (method 1), that would be more rigorous. Thank you.

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    $\begingroup$ I appreciate you trying to work it directly using discriminant, but your method2 is rly clever! $\endgroup$
    – AgentS
    Sep 8, 2019 at 7:57
  • $\begingroup$ @ganeshie8 grace of ganesh hehehe $\endgroup$ Sep 8, 2019 at 8:12

3 Answers 3

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Following your first approach, we have that the discriminant is $$\begin{align} \Delta&=b^2(c−a)^2-4a(b−c)c(a−b)\\ &=b^2(c−a)^2+4ac(b^2-b(a+c)+ac)\\ &=b^2(c+a)^2+(2ac)^2-2b(a+c)(2ac)\\ &=(b(a+c)-2ac)^2 \end{align}.$$ and by letting $\Delta=0$, it is easy to show that $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$.

As regards your second approach, since you already noted that $x=1$ is a root, we have a double root if and only if the following polynomial identity holds (polynomial factorization is unique): $$a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2.$$ Now by letting $x=0$ we find $$c(a-b)=a(b-c)$$ which implies $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$.

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  • $\begingroup$ How did you get that in just one step? I mean... is it an obvious identity or result? $\endgroup$
    – ab123
    Sep 8, 2019 at 7:57
  • $\begingroup$ @ab123 I was actually writing up something more or less along these lines. I didn't bother doing the tedious stuff, I just dumped the quadratic into Wolfram, told it to find the discriminant, and lo and behold one of the alternative forms worked: link $\endgroup$ Sep 8, 2019 at 7:58
  • $\begingroup$ @EeveeTrainer wow! Wolfram is quite intelligent $\endgroup$
    – ab123
    Sep 8, 2019 at 8:01
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    $\begingroup$ @GeorgeCarlin Because $a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2$ is a polynomial identity which holds for any $x$. I edited my answer. $\endgroup$
    – Robert Z
    Sep 8, 2019 at 9:02
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    $\begingroup$ What you wrote is not an identity. The identity is $$ a(b-c)x^2+b(c-a)x+c(a-b)-a(b-c)(x-1)^2=(x-1)(ab+bc-2ac)$$ for all $\,a,b,c,x.\,$ Now if the left side is $0$ and $\,x\ne 1\,$ then we get $\,ab+bc-2ac=0.$ $\endgroup$
    – Somos
    Sep 8, 2019 at 13:08
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If you label: $$\begin{cases}a(b-c)=p\\ b(c-a)=q\\ c(a-b)=r\end{cases} \Rightarrow p+q+r=0$$ then the equation becomes: $$px^2+qx+r=0,\\ D=q^2-4pr=(-p-r)^2-4pr=(p-r)^2=0 \Rightarrow p=r\Rightarrow \\ a(b-c)=c(a-b) \Rightarrow 2ac=ab+bc \stackrel{\cdot \frac1{abc}}{\Rightarrow} \frac2b=\frac1a+\frac1c.$$

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Since $a(b-c)+b(c-a)+c(a-b)=0$, you know that $x=1$ is a root, as you observed.

With synthetic division by $x-1$, we get, assuming $a\ne0$ and $b\ne c$, $$ \begin{array}{c|cc|c} & ab-ac & bc-ab & ac-bc \\ 1 & \downarrow & ab-ac & bc-ac \\ \hline & ab-ac & bc-ac & 0 \end{array} $$ The other root is $$ -\frac{bc-ac}{ab-ac} $$ Therefore $bc-ac=-ab+ac$ is the condition that the equation has two equal roots. This becomes $$ 2ac=bc+ab $$ and, dividing by $abc$, we get $$ \frac{2}{b}=\frac{1}{a}+\frac{1}{c} $$

If $a=0$ or $b=c$ the equation is not a quadratic. Can it be $b=0$? In this case we get $-acx^2+ac=0$ and we need $c\ne0$; the roots are $1$ and $-1$. With $c=0$ it would be the same.

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