Question

If the squared difference of the zeroes of the quadratic polynomial f(x)=x^{​​​​​​2}+px+45 is equal to 144 find the value of p.

Answer 1

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)-(negative of coeff of x)
we can also write ab=45(product of roots)-(constant)
it is given that,
(a - b)² =144
∴ (a + b)² –4ab=144 (calculate)
⇒ (– p)² –4×45=144
⇒ p ²–180=144
⇒ p²= 144+180
⇒p²=324
⇒p= ±√324
∴p=±18