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.

Open in App
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

Suggest Corrections
0