If the sum of the squares of the roots of x2 + px – 3 = 0 is 10, the

1.	If the sum of the squares of the roots of x2 + px – 3 = 0 is 10, then p = A) ± 2 B) ± 3 C) ± 5 D) ± 6
Answer» Correct option is (A) ± 2 Given quadratic equation is \(x^2+px-3=0.\) Let \(\alpha\;and\;\beta\) are roots. \(\therefore\) Sum of roots = -p \(\Rightarrow\) \(\alpha+\beta\) = -p _____________(1) And product of roots = -3 \(\therefore\) \(\alpha\beta\) = -3 _____________(2) Now, \(\alpha^2+\beta^2=10\) (Given) \(\Rightarrow\) \((\alpha+\beta)^2-2\alpha\beta=10\) \(\Rightarrow(-p)^2-2\times-3=10\) \(\Rightarrow p^2\) = 10 - 6 = 4 \(\Rightarrow\) p = \(\pm\,2\) Correct option is A) ± 2

If the sum of the squares of the roots of x2 + px – 3 = 0 is 10, then p = A) ± 2 B) ± 3 C) ± 5 D) ± 6

Answer»

Correct option is (A) ± 2

Given quadratic equation is \(x^2+px-3=0.\)

Let \(\alpha\;and\;\beta\) are roots.

\(\therefore\) Sum of roots = -p

\(\Rightarrow\) \(\alpha+\beta\) = -p _____________(1)

And product of roots = -3

\(\therefore\) \(\alpha\beta\) = -3 _____________(2)

Now, \(\alpha^2+\beta^2=10\) (Given)

\(\Rightarrow\) \((\alpha+\beta)^2-2\alpha\beta=10\)

\(\Rightarrow(-p)^2-2\times-3=10\)

\(\Rightarrow p^2\) = 10 - 6 = 4

\(\Rightarrow\) p = \(\pm\,2\)

Correct option is A) ± 2