# Question 18cf4

Mar 27, 2017

$\left(a\right) : p = \frac{40}{363} , \mathmr{and} , \left(b\right) : \text{ the two roots are, } \frac{2}{33} \mathmr{and} \frac{20}{33.}$

#### Explanation:

We know that, if $\gamma \mathmr{and} \delta$ are the Roots of the

$\gamma + \delta = - \frac{b}{a} , \mathmr{and} , \gamma \delta = \frac{c}{a} \ldots \ldots \ldots \left(\star\right)$

Let the roots of the given quadr. eqn. be, $\alpha \mathmr{and} \beta .$

We are given that, one root is $10$ times the other.

We take, $\beta = 10 \alpha \ldots . . \left(1\right) .$

Applying $\left(\star\right) , \text{ we get, } \alpha + \beta = - \left(- \frac{2}{3}\right) = \frac{2}{3.}$

$\text{Then, } \left(1\right) \Rightarrow \alpha + 10 \alpha = 11 \alpha = \frac{2}{3} , \mathmr{and} , \alpha = \frac{2}{33.}$

Accordingly, $\beta = 10 \alpha = \frac{20}{33.}$

Again, by $\left(\star\right) , \alpha \cdot \beta = \frac{p}{3.}$

$\therefore p = 3 \alpha \cdot \beta = 3 \left(\frac{2}{33}\right) \left(\frac{20}{33}\right) = \frac{40}{363.}$

Thus, $\left(a\right) : p = \frac{40}{363} , \mathmr{and} , \left(b\right) : \text{ the two roots are, } \frac{2}{33} \mathmr{and} \frac{20}{33.}$

Enjoy Maths.!