Three values for quadratic equation?

Question

What is the number of value of $a$ for which the equation $$(a^2-1)x^2-(a^2-3a+2)x+a^2-8a+7=0$$, in $'x'$ possess, three distinct roots?

I got confused over the fact that how does a quadratic equation have 3 distinct roots?

Answer 1

If you have a quadratic equation in $x$ (i.e. $a^2-1\neq0$), then there are exactly 2 solutions (counting with multiplicity).

If you have a linear equation ($a^2-1=0$ and $a^2-3a+2\neq0$), then you have 1 solution.

If the whole thing is zero ($a^2-1=0$, $a^2-3a+2=0$ and $a^2-8a+7=0$), then as people have already said, there are infinitely many solutions, and thus three distinct roots. This is the only way that can happen, so you need to see if you can simultaneously solve these three quadratic equations in $a$, then just state three different numbers for the values of $x$.

Answer 2

If $ux^2+vx+w=0$ is a quadratic equation, then this equation has three distinct root $ \iff u=v=w=0$.

Hence $(a^2-1)x^2-(a^2-3a+2)x+a^2-8a+7=0$ has three distinct root $ \iff a=1$.