Question

❤️❤️hello guys❤️❤️ .... Any intelligent online ..
then solve this ....
best answer will be marked as brainliest ..,

Answer 1

Let the zeroes of the quadratic polynomial be α,β.

Given Quadratic Equation is f(x) = x^2 + px + 45.

Here a = 1, b = p, c = 45.

(i)

We know that sum of zeroes = -b/a

⇒ α + β = -p

(ii)

We know that product of zeroes = c/a

⇒ αβ = 45

Now,

Given that Square of difference of zeroes is equal to 144.

⇒ (α - β)^2 = 144

⇒ α^2 + β^2 - 2αβ = 144

⇒ α^2 + β^2 + (2αβ - 4αβ) = 144

⇒ α^2 + β^2 + 2αβ - 4αβ = 144

⇒ (α + β)^2 - 4αβ = 144

⇒ (-p)^2 - 4(45) = 144

We know that (-p)^2 = -p * -p. We know that -p * -p = +p^2.

⇒ p^2 - 180 = 144

Moving left hand side value to right hand side. Sign changes.

⇒ p^2 = 144 + 180

⇒ p^2 = 324

⇒ p = √324

⇒ p = 18,-18.

Hope this helps!

Answer 2

Hope it helps!!!!!!!!!