Question

the roots of alpha and beta of a quadratic equation x^3-5x+2(k-1)=0 are such that alpha minus alpha beta =1 . Find the value of k​

Answer 1

Answer:

Solution

$x {}^{2} - 5x + 2(k - 1) = 0$

$we \: know \: that \:$

$sum \: of \: roots \: = \frac{ - b}{a} = \frac{ - ( - 5)}{1} = 5$

$product \: of \: roots \: = \frac{c}{a} = \frac{2(k - 1)}{1} = 2(k - 1)$

$\alpha + \beta = 5$

$\alpha - \beta = 1$

$add \: both$

$2 \alpha = 6$

$\alpha = 3$

$substitute \: \alpha \: value \: in \: any \: one \: equation$

$3 - \beta = 1$

$\beta = 3 - 1$

$\beta = 2$

$now \: by \: taking \: product \: of \: the \: roots \:$

$\alpha \times \beta = 2k - 2$

$3 \times 2 = 2k - 2$

$6 = 2k - 2$

$2k = 6 + 2$

$2k = 8$

$k=4$