Question

alpha,beta are the roots of the quadratic equation x²-3x-2=0 then the equation with roots alpha/2,beta/2​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: \alpha , \beta \: are \: the \: roots \: of \: {x}^{2} - 3x - 2 = 0$

We know,

$\boxed{\red{\sf Sum\ of\ the\ roots=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm \implies\: \alpha + \beta = - \dfrac{( - 3)}{1}$

$\rm \implies\:\boxed{ \tt{ \: \alpha + \beta = 3 \: }} - - - (1)$

Also,

$\boxed{\red{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm \implies\: \alpha\beta = \dfrac{ - 2}{1}$

$\rm \implies\:\boxed{ \tt{ \: \alpha\beta = - 2 \: }} - - - (2)$

Now, Consider

$\red{\rm :\longmapsto\:\dfrac{ \alpha }{2} + \dfrac{ \beta }{2} }$

$\rm \: = \: \dfrac{ \alpha + \beta }{2}$

$\rm \: = \: \dfrac{3}{2}$

Now, Consider

$\red{\rm :\longmapsto\:\dfrac{ \alpha }{2} \times \dfrac{ \beta }{2} }$

$\rm \: = \: \dfrac{ \alpha \beta }{4}$

$\rm \: = \: \dfrac{ - 2}{4}$

$\rm \: = \: - \: \dfrac{1}{2}$

So, The required Quadratic equation is

$\rm :\longmapsto\: {x}^{2} - \bigg[\dfrac{ \alpha }{2} + \dfrac{ \beta }{2} \bigg] + \dfrac{ \alpha \beta }{4} = 0$

So, on substituting the values, we get

$\rm :\longmapsto\: {x}^{2} - \dfrac{3}{2} x - \dfrac{1}{2} = 0$

$\rm :\longmapsto\:\dfrac{ {2x}^{2} - 3x - 1}{2} = 0$

$\rm :\longmapsto\: {2x}^{2} - 3x - 1 = 0$

Hence,

$\rm :\longmapsto\: \boxed{ \tt{ \: {2x}^{2} - 3x - 1 = 0 \: }}$

is the required Quadratic equation.