If one of the roots of the equation $(a-b)x^2+ax+1=0$ is double of the other and is real , find the greatest value of $b$.

$\color{green}{a.)\ \dfrac98} \quad \quad \quad \quad \quad b.)\ \dfrac87\\ c.)\ \dfrac86 \quad \quad \quad \quad \quad d.)\ \dfrac75 $

I tried ,

By sum and product of roots formula,

Let one of the root be $\alpha$.

$$\begin{align}3\alpha&=\dfrac{-a}{a-b}\\ \implies 9\alpha^{2}&=\dfrac{a^2}{(a-b)^2}\\ 2\alpha^{2}&=\dfrac{1}{a-b} \end{align}$$

And by discriminant I got $$a^2-4(a-b)\geq 0$$

I look for a short and simple way.

I have studied maths up to $12$th grade.



$$\implies9b=9a-2a^2=-2\left(a-\dfrac94\right)^2+2\left(\dfrac94\right)^2\le2\left(\dfrac94\right)^2$$ as $a$ is real

  • $\begingroup$ how did u get this part from $-2\left(a-\dfrac94\right)^2+2\left(\dfrac94\right)^2\le2\left(\dfrac94\right)^2$ $\endgroup$ – R K Jul 21 '15 at 16:49
  • 1
    $\begingroup$ @RK, As $a-9/4$ is real, $$\left(a-\dfrac94\right)^2\ge0\iff-\left(a-\dfrac94\right)^2\le0$$ $\endgroup$ – lab bhattacharjee Jul 21 '15 at 16:51

$3\alpha=\frac{-a}{a-b}, 2\alpha^2=\frac{1}{a-b}$ divide $\frac{3\alpha}{2\alpha^2}=-a$, $a\alpha=\frac{-3}{2}...............(1)$, As $\alpha$ is a root of $(a-b)x^2+ax+1=0$, $(a-b)\alpha^2+a\alpha+1=0$ $(a-b)\frac{9}{4a^2}-\frac{3}{2}+1=0$ $2a^2-9a+9b=0$ this is quadratic in $a$ and as $a$ being real $81-4(2)(9b)\ge0$ $b\le\frac{9}{8}$


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