Question

If the squared difference of the zeroes of the quadratic polynomial f (x) = x² + px + 45 is equal to 144 , find the value of p .

Answer 1

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

Answer 2

According to the question we know that The given quadratic polynomial is fx = x²+px+45

we can say that α+ β= -p

and also, say αβ = 45

Now,

we can show that

( α + β )² - 4αβ = (α - β )²

(α + β)² - 4αβ = 144

we know that (α+β)² = -p and αβ = 44 then we put it value in given equation

So,

p² - 4 × 45 = 144

p ² - 180 = 144

p² = 144+ 180

p² = 324

p = √324

p = ±18