If $(p^{2} - 2 p + 1) x^{2} - (p^{2} - 3 p + 2) x + p^{2} - 1 = 0$ has more than two roots, then $p =$

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Solution

A quadratic equation $a x^{2} + b x + c = 0$ is said to have more than two roots if $a = b = c = 0$ .
This equation is called identity equation as it satisfied by every value of x
For given equation to have more than 2 roots:
$p^{2} - 2 p + 1 = 0, p^{2} - 3 p + 2 = 0, p^{2} - 1 = 0$
Now,
$p^{2} - 2 p + 1 = 0$
$(p - 1)^{2} = 0$
$p = 1$
Also,
$p^{2} - 3 p + 2 = 0$
$(p - 1) (p - 2) = 0$
$p = 1, p = 2$
And,
$p^{2} - 1 = 0$
$p = 1, p = - 1$
Hence value of p for given equation has more than two roots is 1.

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