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

Hey!

Let α and β are two roots.
Then, α + β = -p..................(i)
and αβ = 45

ATQ,

(α - β)² = 144
Then,(α+β)² - 4αβ = 144
By .....(i) we get,
→ (-p)² - 4 × 45 = 144 
→ p² - 180 = 144
→ p² = 144 + 180
→ 324


Hence p = ±√324 = ±18