Question

For what values of 'a' the equation $\left(a^2-a-2\right)x^2+\left(a^2-4\right)x+a^2-3a+2=0$ will have three solution { more than two solutions}?

Answer 1

We have

${(a^2-a-2)x}^2+(a^2-4)x+(a^2-3a+2)=0$

The above is a quadratic equation and hence will have a maximum of two roots. But we are asked to find the value of ${}^{\prime }{a}^{\prime }$ for which the above expression will yield more than two roots. And this is possible only when the above equation is an identity in $x$, i.e. it will satisfy all values of $x$ and for it to become an identity we need to have coefficient of $x^2=0$ and coefficient of $x=0$ and constant $=0$.

i.e, $a^2-a-2=0⟶\left(1\right)$

$a^2-4=0⟶\left(2\right)$

$a^2-3a+2=0⟶\left(3\right)$

Solving $\left(1\right)$ we get

$a^2-a-2=0$

$\Rightarrow a^2-2a+a-2=0$

$\Rightarrow (a-2)(a+1)=0$ $\therefore$ $a=2$ or $-1$

Solving $\left(2\right)$ we get

$a^2-4=0$

$\Rightarrow a=\pm 2\therefore a=2$ or $-2$

Solving $\left(3\right)$ we get

$a^2-3a+2=0$

$\Rightarrow a^2-2a-a+2=0$

$\Rightarrow (a-2)(a-1)=0$ $\therefore$ $a=2$ or $1$

So the value of ${}^{\prime }{a}^{\prime }$ will be that which is common to all the three conditions in $\left(1\right)$, $\left(2\right)$, $\left(3\right)$.

$\therefore$ $a=2$ (it satisfies all the three equation)

$\therefore$ $a=2$ the equation becomes an identity in $x$ and hence will have more than two roots.