Question

If α and β are the roots of the quadratic equation 3x^2 +kx +8 =0 and α:β=2:3, then find the value of k​

Answer 1

Step-by-step explanation:

Step-by-step explanation:

Given quadratic equation is : 3x² + kx + 8 = 0

and $\alpha :\beta=2:3$

$\frac{\alpha}{\beta}=\frac{2}{3}\\=>\alpha=\frac{2\beta }{3} ------------------(i)$

Now comparing the given quadratic equation with the: ax² + bx + c = 0

We get,

$\alpha+\beta=\frac{-b}{a}\\=>\alpha+\beta=\frac{-k}{3}\:---------------(ii)\\and\:\alpha\beta =\frac{c}{a}=\frac{8}{3}\:--------------(iii)$

Now, putting the value of equation (i) into the equation (iii)

We get,

$\frac{2\beta}{3}*\beta=\frac{8}{3}\\=>\beta^{2}=\frac{8*3}{2*3}\\=>\beta^{2}=4\\=>\beta=\left \{{{+}\atop {-}}2\right.$

putting the value of $\beta$ in the equation (i), we get,

$\alpha=\left\{ {{+}\atop {-}}\right.\frac{2*2}{3}=\left \{{{+} \atop {-}} \right.\frac{4}{3}$

Case 1:

When $\alpha=\frac{4}{3}\:and\:\beta=2,$

Putting the value of $\alpha\:and\:\beta$ into the equation (ii), We get,

$\alpha+\beta=\frac{-k}{3}\\=>\frac{4}{3}+2=\frac{-k}{3}\\=> \frac{10}{3}=\frac{-k}{3}\\=>k=-10$

Case 2:

and when , $\alpha=\frac{-4}{3}\:and\:\beta=-2$

Putting the value of $\alpha\:and\:\beta$ into the equation (ii), We get,

$\alpha+\beta=\frac{-k}{3}\\=>\frac{-4}{3}-2=\frac{-k}{3}\\=> \frac{-10}{3}=\frac{-k}{3}\\=>k=10$

∴ k = -10 or 10

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