For what value of $p$ does $(p^2-16)x^2-(p^2-4p)x+(p^2-5p+4)=0$ have more than $2$ solutions in the variable $x$?

Question

The answer key suggest that solution is (d) $p=4$ but I think that if $p=4$ then it do not remain quadratic equation since for being quadratic equation square coefficient should be non zero but for $p=4$ it is not quadratic equation so please help regarding this.

Answer 1

If we have $p=4$ then the equation becomes $0x^2+0x+0=0$ which is trivial, so unless they were trying so say it has infinitely many solutions. However a quadratic equation can only have either $2$ real roots, one repeated root or no real roots.

Hope this helps!

Answer 2

This question make very little sense in my point of view, but anyway:

We can 'factor' the equation:

$$(p^2-16)x^2-(p^2-4p)x+(p^2-5p+4)=(p-4)[(p+4)x^2-px+p-1]=0$$

If $p=4$, then all values of $x$ are solutions, since the expression becomes zero.

If $p\neq 4$, then we are left with a quadratic equation, which can have at most 2 solutions. (If $p=-4$, then it is strictly speaking a linear equation, which can have at most $1$ solution.)