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Forming quadratic equation - class-XII

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QuestionClick to flip

Sum of roots is $-1$ and sum of their reciprocals is $\dfrac{1}{6}$, then equation is?

AnswerClick to flip back

$x^2+x-6=0$

💡 Explanation:

$\Rightarrow$ Let $\alpha$ and $\beta$ are roots of the equation.

According to the given condition,

$\Rightarrow$ $\alpha+\beta=-1$ ------ ( 1 )

Again according to the given condition,

$\Rightarrow$ $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{1}{6}$

$\Rightarrow$ $\dfrac{\beta+\alpha}{\alpha\beta}=\dfrac{1}{6}$

$\Rightarrow$ $6(\alpha+\beta)=\alpha\beta$

$\Rightarrow$ $6(-1)=\alpha\beta$ [ From ( 1 ) ]

$\therefore$ $\alpha\beta=-6$ ---- ( 2 )

Now, required equation,

$\Rightarrow$ $x^2-(\alpha+\beta)x+(\alpha\beta)=0$

Using ( 1 ) and ( 2 ) we get,

$\Rightarrow$ $x^2-(-1)x+(-6)=0$

$\therefore$ $x^2+x-6=0$