If the difference of the roots of the quadratic equation x2 – ax + b

1.	If the difference of the roots of the quadratic equation x2 – ax + b = 0 is 1, then A) a2 – 4b = -1 B) a2 – 4b = 4 C) a2 – 4b = 1 D) a2 – 4b = 0
Answer» Correct option is (C) a²– 4b = 1 Let \(\alpha\;and\;\beta\) are roots of quadratic equation \(x^2-ax+b=0.\) \(\therefore\) Sum of roots \(=\frac{-(-a)}1=a\) \(\Rightarrow\) \(\alpha+\beta=a\) ______________(1) & Product of roots \(=\frac b1=b\) \(\Rightarrow\) \(\alpha\beta=b\) ______________(2) Now, \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\) \(\Rightarrow\) \((\alpha-\beta)^2=a^2-4b\) ______________(3) Given that difference of the roots of given equation is 1. \(\therefore\) \(\alpha-\beta=1\) Then from (3), we have \(a^2-4b=1^2=1\) \(\Rightarrow\) \(a^2-4b=1\) Correct option is C) a² – 4b = 1

If the difference of the roots of the quadratic equation x2 – ax + b = 0 is 1, then A) a2 – 4b = -1 B) a2 – 4b = 4 C) a2 – 4b = 1 D) a2 – 4b = 0

Answer»

Correct option is (C) a²– 4b = 1

Let \(\alpha\;and\;\beta\) are roots of quadratic equation \(x^2-ax+b=0.\)

\(\therefore\) Sum of roots \(=\frac{-(-a)}1=a\)

\(\Rightarrow\) \(\alpha+\beta=a\) ______________(1)

& Product of roots \(=\frac b1=b\)

\(\Rightarrow\) \(\alpha\beta=b\) ______________(2)

Now, \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\)

\(\Rightarrow\) \((\alpha-\beta)^2=a^2-4b\) ______________(3)

Given that difference of the roots of given equation is 1.

\(\therefore\) \(\alpha-\beta=1\)

Then from (3), we have

\(a^2-4b=1^2=1\)

\(\Rightarrow\) \(a^2-4b=1\)

Correct option is C) a² – 4b = 1