Question

if alpha and beta are the roots of the equation 3x^2-6x+1=0 then find the quadratic equation whose roots are 2alpha +beta and 2beta +alpha​

Answer 1

Answer:

From eq.,

$3 {x}^{2} - 6x + 1 = 0$

$\alpha + \beta = - \frac{ - 6}{3} = 2$

$\alpha \beta = \frac{1}{3}$

now, sum of the roots of new eq. is

$2 \alpha + \beta + 2 \beta + \alpha = 3( \alpha + \beta ) = 3 \times 2 = 6$

product of roots of new eq is

$(2 \alpha + \beta )(2 \beta + \alpha ) = 4 \alpha \beta + 2 { \beta }^{2} + 2 { \alpha }^{2} + \alpha \beta =$

$5 \alpha \beta + 2( { \alpha }^{2} + { \beta }^{2} ) = 5 \alpha \beta + 2(( \alpha + \beta ) {}^{2} - 2 \alpha \beta ) = 5 \times \frac{1}{3} + 2( {2}^{2} - 2 \times \frac{1}{3} ) = \frac{5}{3} + 2(4 - \frac{2}{3} ) =$

$\frac{5}{3} + \frac{20}{3} = \frac{25}{3}$

so quadratic equation is,

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

${x}^{2} - 6x + \frac{25}{3} = 0$

$3 {x}^{2} - 18x + 25 = 0$

Ans.