Question

If alpha, beta are the roots of $2{{x}^{2}}+x+3=0$ then $\dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }$ is?

Answer 1

Hint:Here we have to find the value of the expression given by using the quadratic equation given. Firstly we will write the sum and product rule of roots of a quadratic equation and get a relation between alpha and beta. Then using our quadratic equation we will get two equations and two values. Finally we will simplify the expression given by taking LCM and substitute those values in it and get our desired answer.

Complete step by step answer:
We have been given that the two roots of the below equation is alpha and beta:
$2{{x}^{2}}+x+3=0$…$\left( 1 \right)$
The two roots are $\alpha ,\beta $ .
We have to find the value of:
$\dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }$….$\left( 2 \right)$
Now we know that the for a general form quadratic equation $a{{x}^{2}}+bx+c=0$ whose two roots are $\alpha ,\beta $ the sum and product of these two roots are given as:
$\alpha +\beta =-\dfrac{b}{a},\alpha .\beta =\dfrac{c}{a}$ ….$\left( 3 \right)$
On comparing the general form by equation (1) we get,
$a=2,b=1,c=3$
On substituting the above value in equation (3) we get,
$\alpha +\beta =-\dfrac{1}{2},\alpha .\beta =\dfrac{3}{2}$…$\left( 4 \right)$
Next we will simplify the value in equation (2) by taking the LCM as follows:
$\Rightarrow \dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }=\dfrac{\left( 1-\alpha \right)\left( 1+\beta \right)+\left( 1-\beta \right)\left( 1+\alpha \right)}{\left( 1+\alpha \right)\left( 1+\beta \right)}$
$\Rightarrow \dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }=\dfrac{\left( 1\times 1+1\times \beta -\alpha \times 1-\alpha \times \beta \right)+\left( 1\times 1+1\times \alpha -\beta \times 1-\beta \times \alpha \right)}{1\times 1+1\times \beta +\alpha \times 1+\alpha \times \beta }$
Simplifying further we get,
$\Rightarrow \dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }=\dfrac{\left( 1+\beta -\alpha -\alpha \beta \right)+\left( 1+\alpha -\beta -\alpha \beta \right)}{1+\beta +\alpha +\alpha \beta }$
$\Rightarrow \dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }=\dfrac{2-2\alpha \beta }{1+\left( \alpha +\beta \right)+\alpha \beta }$
Substitute the value from equation (4) above,
$\Rightarrow \dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }=\dfrac{2-2\times \dfrac{3}{2}}{1-\dfrac{1}{2}+\dfrac{3}{2}}$
$\Rightarrow \dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }=\dfrac{2-3}{\dfrac{2-1+3}{2}}$
Simplifying further,
$\Rightarrow \dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }=\dfrac{-1}{\dfrac{4}{2}}$
$\Rightarrow \dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }=-\dfrac{1}{2}$

Hence if alpha, beta are the roots of $2{{x}^{2}}+x+3=0$ then $\dfrac{1-\alpha }{1+\alpha }+\dfrac{1-\beta }{1+\beta }$ is equal to $-\dfrac{1}{2}$.

Note:Quadratic equations are those equations whose unknown variable has the highest power as $2$ . A quadratic equation has two roots. We have used the relation between the roots and the coefficient of a quadratic equation in this equation for finding the value given. As we have to find the value of an expression with fraction values, taking the LCM and simplifying them should be our first step as we have the value of roots in sum and product form.