If the squared difference of the zeros of polynomial $x^{2} + P C + 45 = 144$ then find p.

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Solution

Let $α, β$ are the roots of the quadratic polynomial $f (x) = x^{2} + p x + 45$ then
$α + β = - p - - - - (1)$ and $α β = 45$
Given $(α - β)^{2} = 144$
$∴ (α + β)^{2} - 4 α β = 144$
$\Rightarrow (- p)^{2} - 4 \times 45 = 144$ [Using (1)]
$\Rightarrow p^{2} - 180 = 144 \Rightarrow p^{2} = 144 + 180 = 324 \Rightarrow p = \pm \sqrt{324} = \pm 18$
Thus, the value of p is $\pm 18.$

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