Question

If the square of the difference of zeroes of quadratic polynomial f(x) =x^2+px+45 is equal to 144. Find P. ​

Answer 1

Step-by-step explanation:

A=alpha,b=beta

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