If the Squared Difference of the Zeros of Polynomial x2+px+45 is 144 Then Find p

If the Squared Difference of the Zeros of Polynomial x²+px+45 is 144 Then Find p

Given the polynomial x^2 + px + 45, and the squared difference of the zeros is 144.

We know that the squared difference of the zeros (α - β)^2 is equal to (α + β)^2 - 4αβ.

Using the formula for the sum of the zeros (α + β) and the product of the zeros αβ, we have:

(α - β)^2 = (α + β)^2 - 4αβ

Substituting the values into the equation, we get:

144 = (-p)^2 - 4(45)

Simplifying further:

144 = p^2 - 180

Rearranging the equation:

p^2 = 324

Taking the square root of both sides:

p = ±√324

p = ±18

Therefore, the value of p can be either 18 or -18, depending on the positive or negative square root.

Hence, the possible values of p are 18 and -18.

Answer

The value of p are 18 and -18.

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