Question

if alpha and beta are roots of quadratic equation 3x2+kx+8 and alpha/beta=2/3 then find the value of k​

Answer 1

ANSWER:-

Given:

If α and β are the roots of quadratic equation 3x² +kx+8 & α/β=2/3.

To find:

The value of k.

Explanation:

The given quadratic equation as compared with Ax² +Bx+C.

Therefore,

• A= 3
• B= k
• C= 8

&

α and β are the roots of the above equation.

Given,

$\frac{\alpha }{\beta } =\frac{2}{3}$

$\alpha =\frac{2\beta }{3}$

Now,

Sum of the zeroes:

α+β= $-\frac{coefficient\:\:of\:x}{coefficient\:\:of\:x^{2} }$

Or

α+β= $-\frac{b}{a}$

α+β= $-\frac{k}{3}$

$=\:\frac{2\beta }{3} +\beta =-\frac{k}{3} \\\\=\:\frac{2\beta +3\beta }{3} =-\frac{k}{3} \\\\=\:\frac{5\beta }{3} =-\frac{k}{3} \\\\=\:5\beta =-k\\\\=\:\beta = -\frac{k}{5}$

Therefore,

Putting the value of β in α value:

α= $=\:\frac{2\beta }{3} =\frac{2}{3} *(-\frac{k}{5} )\\\\=\:\alpha = -\frac{2k}{15}$

Product of the zeroes:

αβ= $\frac{constant\:term}{coefficient\:of\:x^{2} } =\frac{c}{a}$

We have value ,

• α= -2k/15
• β= -k/5

So,

$=\:\frac{-2k}{15} \:*\:(\frac{-k}{5} )=\frac{8}{3} \\\\=\:\frac{2k^{2} }{75} =\frac{8}{3} \\\\=\:6k^{2} =\:600\\\\=\:k^{2} =\frac{600}{6} \\\\=\:k^{2} =100\\\\=\:k=\sqrt{100} \\\\=\:k= 10$

Thus,

The value of k is 10.