Question

Solve the equation 18x³+81x²+lambda x +60=0,one root being half the sum of the other two. Hence find the value of lambda.

Answer 1

Final Answer : $\lambda = 121$

Steps:
1) Let the roots be $\alpha,\beta and \gamma$
We know that,
$\alpha + \beta +\gamma = \frac {-81} {18} = \frac{-9}{2} --(1)$

Then , according to the question we have,
$\alpha = \frac {\beta +\gamma } {2} \\ => \alpha = \frac{1}{2}.(\frac{-9}{2}-\alpha) \\ => \alpha = \frac{-3}{2}$

2) We got, one root of the equation as $\frac{-3}{2}$

So,
$18 \alpha^3 +81\alpha ^2 +\lambda x +60=0 \\ =>18 \frac{(-3)}{2}^3 +81 \frac{(-3)}{2}^2+ \lambda \frac{(-3)}{2}+ 60 =0 \\ => \lambda = 121$

Therefore, Required value is $121$

Answer 2

In the attachments I have answered this problem .

Let the roots be alpha, beta, gamma

As per given data

Alpha = ( beta + gamma) /2


Concept:

S1 = sum of the roots taken one at

a time.

See the attachment for detailed solution.