If the sum of the roots of the quadratic equations is 1 and sum of the squares of the roots is 13 , then…
If the sum of the roots of the quadratic equations is 1 and sum of the squares of the roots is 13 , then find that equation.
\(x^2+x-6=0\)
\(x^2-x+6=0\)
\(x^2-x-6=0\)
\(x^2+x+6=0\)
Solution
Let the roots of the quadratic equation are \(\alpha\) and \(\beta\) then it is given that
\(\begin{aligned}
\alpha+\beta & =1 \text { and } \alpha^2+\beta^2=13 \\
\therefore \quad \alpha \beta & =\frac{1}{2}\left[(\alpha+\beta)^2-\left(\alpha^2+\beta^2\right)\right]=\frac{1}{2}[1-13]=-6
\end{aligned}\)
So, equation of required quadratic is
\(x^2-(\alpha+\beta) x+\alpha \beta=0 \Rightarrow x^2-x-6=0.\)