Brainlyif the squared difference of the zeros of the quadratic polynomial f(x) = x2 +px +45 is equal to 144, find the value of p.
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Hey
Here is your answer,
Given quadratic polynomial f(x) = x2+px+45
Let α and β be the roots(zeros) of the given quadratic polynomial f(x) = x2+px+45
Also, given (α - β)2 = 144
We know that, (α - β)2 = (α + β)2 - 4αβ
Now, we have from the quadratic polynomial f(x) = x2+px+45, (α + β) = -p, αβ = 45
⇒144 = (-p)2 - 4(45)
⇒144 = p2 - 180
⇒144 + 180 = p2
⇒324 = p2
⇒p2 = 324
⇒p = 18
Hope it helps you!
Here is your answer,
Given quadratic polynomial f(x) = x2+px+45
Let α and β be the roots(zeros) of the given quadratic polynomial f(x) = x2+px+45
Also, given (α - β)2 = 144
We know that, (α - β)2 = (α + β)2 - 4αβ
Now, we have from the quadratic polynomial f(x) = x2+px+45, (α + β) = -p, αβ = 45
⇒144 = (-p)2 - 4(45)
⇒144 = p2 - 180
⇒144 + 180 = p2
⇒324 = p2
⇒p2 = 324
⇒p = 18
Hope it helps you!
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