Question

(i) Solve the equation $24x^3-14x^2-63x+\lambda =0$, one root being double of another. Hence find the value(s) of $\lambda$.
(ii) Solve the equation $18x^3+81x^2+\lambda x+60=0$, one root being half the sum of the other two. Hence find the value of $\lambda$.

Answer 1

Given 1 root is double of another

$24x^2-14x^2-63x+\lambda =0...\left(1\right)$

Let $\alpha ,\beta ,\gamma$ be the roots of eq (1) then using property

$\alpha +\beta +\gamma =\frac{14}{24}\Rightarrow 3\alpha +\gamma =\frac{14}{24}...\left(1\right)$ $[\beta =2\alpha ]$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{-63}{24}.\Rightarrow 2{\alpha }^2+3\alpha \gamma =\frac{-63}{24}...\left(2\right)$

$\alpha \beta \gamma =\frac{\lambda }{24}\Rightarrow {\alpha }^2\gamma =\frac{-\lambda }{24}...\left(3\right)$

form eq (1) & (2) we get $2{\alpha }^2+3\alpha (\frac{14}{24}-3\alpha )=\frac{-63}{24}$

$168{\alpha }^2-42\alpha -63=0$

$8{\alpha }^2-2\alpha -3=0\Rightarrow (\alpha =\frac{3}{4},\frac{-1}{2})$

For $\alpha =\frac{3}{4}\gamma =\frac{-5}{3}$ [ from eq (1)]

For $\alpha =\frac{-1}{2}\gamma =\frac{-3}{4}$

Hence, $\lambda =-48{\alpha }^2\gamma$ [from eq 3]

Then $[\lambda =45\lambda =18]$

$(\alpha =\frac{3}{4},\gamma =-\frac{5}{3})$

$(\alpha =\frac{-1}{2}\gamma =\frac{-3}{4})$

(ii) Given $18x^3+81x^2+\lambda x+60=0,...\left(1\right)$ 1 root is half the sum of other two

Let $\alpha ,\beta ,\gamma$ be root of eq (1)

then $\alpha +\beta +\gamma =\frac{-81}{18}\Rightarrow 3\alpha =\frac{-81}{18}\Rightarrow [\alpha =\frac{-3}{2}][\because \alpha =\frac{\beta +\gamma }{2}]$

$\alpha \beta \gamma =\frac{-60}{18}$

$\beta \gamma =\frac{-66}{18},\frac{-2}{3}=\frac{20}{9}$

then $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{\lambda }{18}.\Rightarrow \alpha (\beta +\gamma )+\beta \gamma =\frac{\lambda }{18}$

$\lambda =18\left[\alpha \right(\beta +\gamma )+\beta \gamma ]=18[\frac{-3}{2}\times -3+\frac{20}{9}]$

$\lambda =18\left[\frac{81+40}{18}\right]=121.$

$[\lambda =121]$

Hence, value of $\lambda$ is 121.