Question

In the equation $x^2-\frac{10x}{9}+c=0$, one solution is the square of the other solution. If $c>0$ is rational number $\left(\frac{m}{n}\right)$ in the lowest term, then the value of $(m+n)$ is equal to

• A. 25
• B. 28
• C. 32
• D. 35

Answer 1

Answer: (D)

Let the roots are $\alpha$ and ${\alpha }^2$
$\therefore x^2-\frac{10x}{9}+c=(x-\alpha )\left(x-{\alpha }^2\right)$
$\alpha +{\alpha }^2=\frac{10}{9}\text{ and }\alpha \cdot {\alpha }^2=c$
$9{\alpha }^2+9\alpha -10=0\Rightarrow 9{\alpha }^2+15\alpha -6\alpha -10=0\Rightarrow 3\alpha (3\alpha +5)-2(3\alpha +5)=0$
$\therefore \alpha =\frac{2}{3}\text{ or }\alpha =\frac{-5}{3}$
as $c>0$ and ${\alpha }^3=c$, hence $\alpha =\frac{-5}{3}$ is rejected
$\therefore \alpha =\frac{2}{3};c={\alpha }^3=\frac{8}{27}$
$\Rightarrow (m+n)=35$