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Question

If the squared difference of the zeros of polynomial x2+PC+45=144then find p.

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Solution

Let α,β are the roots of the quadratic polynomial f(x)=x2+px+45 then
α+β=p(1) and αβ=45
Given (αβ)2=144
(α+β)24αβ=144
(p)24×45=144 [Using (1)]
p2180=144p2=144+180=324p=±324=±18
Thus, the value of p is ±18.

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