2
$\begingroup$

If both roots of the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that $2a=b+c$.

Things should be known:

  • Roots of a Quadratic Equations can be identified by:

    The roots can be figured out by: $$\frac{-b \pm \sqrt{d}}{2a},$$ where $$d=b^2-4ac.$$

  • When the equation has equal roots, then $d=b^2-4ac=0$.

  • That means $d=(b-c)^2-4(a-b)(c-a)=0$

$\endgroup$
  • $\begingroup$ What is $(b-c)^2 + 4bc$? $\endgroup$ – Alex B. Jan 4 '13 at 15:22
  • 3
    $\begingroup$ So have you tried working with your expression for $d$? $\endgroup$ – Mark Bennet Jan 4 '13 at 15:22
  • $\begingroup$ Yes, I tried to evaluate the $d$, It was (not completely evaluated) $$d= (b-c)^2-4(a-b)(c-a)=0$$ Is it right? And after this I am not able to understand that how to proceed now. Can you help? $\endgroup$ – Saharsh Jan 4 '13 at 15:27
  • $\begingroup$ Expand $(-2a+b+c)^2$ . Do you see some similarity? $\endgroup$ – Inquisitive Jan 4 '13 at 15:48
2
$\begingroup$

As the two roots are equal the discriminant must be equal to $0$.

$$(b-c)^2-4(a-b)(c-a)=(a-b+c-a)^2-4(a-b)(c-a)=\{a-b-(c-a)\}^2=(2a-b-c)^2=0 \iff 2a-b-c=0$$


Alternatively, solving for $x,$ we get $$x=\frac{-(b-c)\pm\sqrt{(b-c)^2-4(a-b)(c-a)}}{2(a-b)}=\frac{c-b\pm(2a-b-c)}{2(a-b)}=\frac{c-a}{a-b}, 1$$ as $a-b\ne 0$ as $a=b$ would make the equation linear.

So, $$\frac{c-a}{a-b}=1\implies c-a=a-b\implies 2a=b+c$$

$\endgroup$
4
$\begingroup$

The expression for $d$ factors as $(2a-b-c)^2$, so we must have $2a-b-c=0$.

The easiest way to see this is to let $x=a-b,y=c-a$ and note that $b-c=-x-y$: $d=(-x-y)^2-4xy=x^2+y^2+2xy-4xy=(x-y)^2$.

Alternative solution: Note that $(a-b)+(b-c)+(c-a)=0$, i.e. $x=1$ is a root of the equation. If both roots are equal, it means that the other root is also $1$, showing that $(a-b)x^2+(b-c)x+(c-a)=(a-b)(x-1)^2$.

Equating the constant term, we find: $c-a=a-b$, which means $2a=b+c$. (This can be done with the coefficient of $x$ too).

$\endgroup$
1
$\begingroup$

Edit: $(b-c)^2=4(a-b)(c-a)$, so $a-b$ and $c-a$ are either both nonnegative or both nonpositive. In the former case, put $x=a-b$ and $y=c-a$; in the latter case, put $x=-(a-b)$ and $y=-(c-a)$. Now the A.M. $\ge$ G.M. inequality says that $$ \frac{x+y}2\ge\sqrt{xy},\ \text{ i.e. } (x+y)^2\ge4xy $$ for $x,y\ge0$, and equality holds if and only if $x=y$.

$\endgroup$
  • 2
    $\begingroup$ Sorry, but I didn't get the answer. Can you clarify what's A.M. and G.M. $\endgroup$ – Saharsh Jan 4 '13 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.