Question

If squared difference of the zeros of the quadratic polynomial x²+px+45 is equal to 144 , find the value of p. *

Answer 1

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Answer 2

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If squared difference of the zeros of the quadratic polynomial x²+px+45 is equal to 144 , find the value of p. *

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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