Question

If $\alpha$ and $\beta$ are the zeroes of the equation $6x^2+x-2=0$, find $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

Answer 1

Consider the given equation.

6x2+x−2=06x2+x−2=0 …….. (1)

Since, α,βα,β are roots of given equation.

So,

α+β=−ba=−16α+β=−ba=−16

αβ=ca=−26=−13αβ=ca=−26=−13

Since,

=αβ+βα=αβ+βα

=α2+β2αβ=α2+β2αβ

=(α+β)2−2αβαβ=(α+β)2−2αβαβ

=(−16)2−2×−13−13=(−16)2−2×−13−13 =\dfrac{{{\left( -\dfrac{1}{6} \right)}^{2}}-2\times -\dfrac{1}{3}}{-\dfrac{1}{3}}

=136+23−13=136+23−13

=1+2436−13=1+2436−13

=−2512

Hence, the value is −2512−2512.