Question

solve this
anyone Interested

Answer 1

Let the roots of the
${x}^{2} - x + k = 0$
be
$\alpha \: and \: \beta$
And roots of
${x}^{2} - x + 3k = 0$
be
$2 \alpha \: and \: \gamma$
Using the sum and product of roots relation in 1st equation,

$\alpha + \beta = 1 \\ \alpha \beta = k$

Using the sum and product of roots relation in 2nd equation,

$2 \alpha + \gamma = 1 \\ 2 \alpha \gamma = k$

Solving these four equations simultaneously, We get the Following solutions,

$( \alpha , \: \beta , \: \gamma , \: k) = (0, \: 1, \: 1, \: 0)$
Or

$( \alpha , \: \beta , \: \gamma , \: k) = ( - 1, \: 2, \: 3, \: - 2)$
It is observable that the second set of solutions is correct as we know that one root of an equation is double of a root of another equation, If the root would have been 0, then its double would be 0 as well, so we can say that 1st set of solutions is not valid.

Thus, from the second set,
the value of k = -2.

Answer 2

sorry don't know bro