Question

If the squared difference of the zeros of the quadratic polynomial $f(x)=x^{2}+px+45$ is equal to 144, find the value of p.

Answer 1

SOLUTION :

Let the two zeroes of the polynomial be α and β.

Given :  f(x)= x² + px + 45

On comparing with ax² + bx + c,

a = 1 , b = p , c = 45  

Sum of the zeroes = −coefficient of x / coefficient of x²  

α + β  = -b/a = -p/1 = -p

α+β = - p ……………………….(1)

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = 45/1 = 45

α×β =  45 ....................................(2)

(α - β)² =144        [Given]

As we know that ,  α - β)² = α² + β² - 2αβ

α² + β² - 2αβ = 144

As we know that  , α² + β² = (α + β)² - 2αβ

(α + β)² - 2αβ - 2αβ = 144

(α + β)² - 4αβ = 144

By Substituting the value from eq 1 & 2

p² - 4 × 45 =144

p² - 180 = 144

p² = 144 + 180

p² = 324

p = √324

p = ±18

Hence, the value of p is  ±18

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Answer 2

Answer:

SOLUTION :

-p/1 = -p

If the squared difference of the zeros of the quadratic polynomial $f(x)=x^{2}+px+45$ is equal to 144, find