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Question

If the sqaure difference of the zeros of the quadratic polynomial

F(x)= x^2 +px+45 is equal to 144, find the value of p.

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Solution

we can let ' a ' and ' b ' be the roots of the quadratic polynomial f (x) = x² + px + 45
∴ we can write a+b= -p (sum of the roots)
we can also write ab=45(product of roots)
it is given that,
∴ (a + b)² –4ab=144
⇒ (– p)² –4×45=144
⇒ p ²–180=144
⇒ p²= 144+180
⇒p²=324
⇒p= ±√324
∴p=±18

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