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.

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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