Question

if if alpha and beta are the roots of quadratic equation 3 X square + kx + 8 is equal to zero and alpha upon beta equal to 3 then find the value of k ​

Answer 1

Answer:

$k = ±8 \sqrt{2}$

Step-by-step explanation:

We know that standard Quadratic equation

$a {x}^{2} + bx + c = 0$

it has two roots,let the roots are

$\alpha\:and\: \beta$

We know that relation between coefficient and zeros are

$\alpha + \beta = \frac{ - b}{a} \\ \\ \alpha \beta = \frac{c}{a}$

Here Quadratic equation is

$3 {x}^{2} + kx + 8 = 0 \\ \\ \alpha + \beta = \frac{ - k}{3}...eq1 \\ \\ \alpha \beta = \frac{8}{3}...eq2$

Given that

$\frac{ \alpha }{ \beta } = 3...eq3$

From eq3,put the value of alpha in eq1

$3 \beta + \beta = \frac{ - k}{3} \\ \\ 4 \beta = \frac{ - k}{3} \\ \\ \beta = \frac{ - k}{12}$

put value of beta in eq2

So

$\alpha ( \frac{ - k}{12} ) = \frac{8}{3} \\ \\ \alpha = \frac{ - 32}{k}$

put the value of alpha and beta in eq1

So

$\frac{ - 32}{k} + \frac{ - k}{12} = \frac{ - k}{3} \\ \\ \frac{ - 384 - {k}^{2} }{12k} = \frac{ - k}{3} \\ \\ 3(384 + {k}^{2} ) = 12 {k}^{2} \\ \\ 384 + {k}^{2} = 4 {k}^{2} \\ \\ 384 = 3 {k}^{2} \\ \\ {k}^{2} = \frac{384}{3} \\ \\ {k}^{2} = 128 \\ \\ k = ± 8 \sqrt{2}$

Hope it helps you.