Question

In the equation $x^2-\frac{10 x}{9}+c=0$, 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{10 x }{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} $
$\Rightarrow( m + n )=35$