Question

What will be the value of (α – β)^2, if α and β are the zeros of 4x^2 – 27x – 40?

(a) \(\frac {1369}{16}\)

(b) \(\frac {139}{16}\)

(c) \(\frac {1369}{6}\)

(d) \(\frac {19}{16}\)

I had been asked this question by my college director while I was bunking the class.

My doubt is from Zeros and Coefficients of Polynomial in chapter Polynomials of Mathematics – Class 10

Answer 1

Correct answer is (a) \(\frac {1369}{16}\)

Easiest explanation: α and β are the zeros of 4x^2 – 27x – 40

α + β = \(\frac {-27}{4}\) and αβ = – 10

(α – β)^2 = α^2 + β^2 – 2αβ

(α – β)^2 = (α + β)^2 – 2αβ – 2αβ

(α – β)^2 = \((\frac {-27}{4})\)^2 – 4(-10)

(α – β)^2 =  \(\frac {729}{16}\) + 40 =  \(\frac {1369}{16}\)