Question

if alpha and bita are the roots of the following equation 5x^2-3x-2=0 then find the equation whose roots are alpha+2bita and 2alpha+bita​

Answer 1

Answer:

$25(\frac{25x^{2}-45x+8}{25})$

Step-by-step explanation:

$f(x) = 5x^{2} - 3x- 2$

Given,

$5x^{2}-3x-2=0\\or,\\f(x)=0$

Factorising f(x),

$5x^{2} -3x-2 \\= 5x^{2}-5x+2x-2\\= 5x(x-1)+2(x-1)\\=(5x+2)(x-1)$

Therefore,

$(5x+2)(x-1)=0\\5x+2=0 \\=> x= \frac{-2}{5}\\x-1 =0\\=>x= 1$

$\alpha = \frac{-2}{5}\\\beta = 1$

The new equation has roots $\alpha + 2\beta and \beta + 2\alpha$

$\alpha + 2\beta = \frac{-2}{5} + 2 = \frac{8}{5}\\\\\beta + 2\alpha = 1 + \frac{-4}{5} = \frac{1}{5}\\\\sum = (\frac{8}{5}+ \frac{1}{5}) = \frac{9}{5}\\\\product = (\frac{8}{5} X \frac{1}{5}) = \frac{8}{25}$

The quadratic equation is given by,

$x^{2} - (\alpha + \beta )x+\alpha \beta$

The quadratic equation is

$x^{2}-\frac{9}{5}x+\frac{8}{25}\\=\frac{25x^{2}-45x+8}{25}$

The equation is

$25(\frac{25x^{2}-45x+8}{25})$