Question

Suppose the roots of the equation 2x^2−5x−6=0 are α and β.Find the quadratic equation with roots 1/α and 1/β.​

Answer 1

The quadratic equation with roots 1/α and 1/β is $6 x^{2}+5 x-2=0$

Solution:

Given, roots of equation $2 x^{2}-5 x-6=0 \text { are } \alpha, \beta$

We have to find equation whose roots are 1/α, and 1/β.

We know that,

$\text { sum of roots }=\frac{-x \text { cosficient }}{x^{2} \text { coefficient }} \rightarrow \alpha+\beta=\frac{-(-5)}{2} \rightarrow \alpha+\beta=\frac{5}{2}$

$\text { And, product of roots }=\frac{c o n s t a n t}{x^{2} c o e f f i c i e n t} \rightarrow \alpha \beta=\frac{-6}{2} \rightarrow \alpha \beta=-3$

Now, we know that, general form of an equation is

$x^{2}-(\text { sum of roots }) x+\text { product of roots }=0$

Then, equation whose roots 1/α and 1/β is

$\begin{array}{l}{x^{2}-\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) x+\frac{1}{\alpha} \times \frac{1}{\beta}=0} \\\\ {\rightarrow x^{2}-\left(\frac{\alpha+\beta}{\alpha \beta}\right) x+\frac{1}{\alpha \beta}=0}\end{array}$

from above given equation,

$\begin{array}{l}{\rightarrow x^{2}-\left(\frac{\frac{5}{2}}{-3}\right) x+\frac{1}{-3}=0} \\\\ {\text { multiplying equation with }-3} \\\\ {\rightarrow-3 x^{2}-\frac{5}{2} x+1=0} \\\\ {\text { multiplying equation with } 2} \\\\ {\rightarrow-6 x^{2}-5 x+2=0} \\\\ {\rightarrow-6 x^{2}-5 x+2=0} \\\\ {\text { multiplying equation with }-1} \\\\ {\rightarrow 6 x^{2}+5 x-2=0} \\\\ {\text { Hence, the required line equation is } 6 x^{2}+5 x-2=0}\end{array}$