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

What will be the value of (α – β)^2, if α and β are the zeros of

1.	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» 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 =\(\frac {1369}{16}\)

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