Question

If α β are the roots of equation 2x(2x+1) = 1 then β =?

Answer 1

Step-by-step explanation:

please refer to the attached file

Answer 2

ANSWER.

Option [ 2 ] is correct option.

$\sf \implies \: - 2 \alpha ( \alpha + 1)$

$\sf \to \: \green{{ \underline{step - \: by - \: step \: - explanation}}}$

$\sf \to \: \alpha \: and \: \beta \: \: are \: the \: roots \: of \: the \: equation \: 2x(2x + 1) = 1 \\ \\ \sf \to \: 4 {x}^{2} + 2x - 1 = 0 \\ \\ \sf \to \: sum \: of \: the \: roots \: of \: the \: equation \: ( \alpha + \beta ) = \dfrac{ - b}{a} \\ \\ \sf \to \: \alpha + \beta = \frac{ - 1}{2} \\ \\ \sf \to \: \beta = \frac{ - 1}{2} - \alpha ......(1)$

$\sf \to \: as \: we \: know \: that \: \alpha \: also \: satisfying \: the \: roots \\ \\ \sf \to \: 4 \alpha {}^{2} + 2 \alpha - 1 = 0 \\ \\ \sf \to \: 4 {a}^{2} + 2 \alpha = 1 \\ \\ \sf \to \: divide \: lhs \: \: and \: \: rhs \: \: by \: 2 \\ \\ \sf \to \: \frac{4 \alpha {}^{2} + 2 \alpha }{ \cancel{2}} = \frac{1}{ \cancel{2}} \\ \\ \sf \to \: {2 \alpha }^{2} + \alpha = \frac{1}{2}$

$\sf \to \: put \: the \: value \: of \: \frac{1}{2} \: \: in \: \: equation \: (1) \\ \\ \sf \to \: \beta = - (2 { \alpha }^{2} + \alpha ) - \alpha \\ \\ \sf \to \: \beta = - 2 \alpha {}^{2} - \alpha - \alpha \\ \\ \sf \to \: \beta = - 2 \alpha ( \alpha + 1) = answer$