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

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Solution

Given equation is $(p^{2} - 2 p + 1) x^{2} - (p^{2} - 3 p + 2) x + (p^{2} - 1) = 0$

$p^{2} - 3 p + 2$
$= p^{2} - 2 p - p + 2$
$= p (p - 2) - 1 (p - 2)$
$= (p - 2) (p - 1)$

Now,
$(p^{2} - 2 p + 1) x^{2} - (p^{2} - 3 p + 2) x + (p^{2} - 1) = 0$
$\Rightarrow (p - 1)^{2} x^{2} - (p - 1) (p - 2) x + (p - 1) (p + 1) = 0$
$\Rightarrow (p - 1) [(p - 1) x^{2} - (p - 2) x + (p + 1)] = 0$

Given $(p^{2} - 2 p + 1) x^{2} - (p^{2} - 3 p + 2) x + (p^{2} - 1) = 0$ has more than two roots
since a quadratic equation has maximum of two roots So it means the above equation should be correct for all values of $x$

It is possible only if, $p - 1 = 0 ⟹ p = 1$

Hence, the value of $p$ is $1$

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