Question

If the two roots of $9x^3+24x^2+13x+2=0$ are equal, then the roots are:

• A. $-2,\frac{1}{3},\frac{1}{3}$
• B. $2,-\frac{1}{3},-\frac{1}{3}$
• C. $1,-\frac{1}{3},-\frac{1}{3}$
• D. $-2,-\frac{1}{3},-\frac{1}{3}$

Answer 1

The correct option is D $-2,-\frac{1}{3},-\frac{1}{3}$

Let the roots of the equation be $a,a,b$.
$\therefore$ the sum of roots is $2a+b=-\frac{24}{9}$....... (i)
The product of roots: $a^{2}b=-\frac{2}{9}$.............(ii)
Product of roots taken two at a time: $a(a+2b)=\frac{13}{9}$........(iii)
$\because a\ne 0$
(iii) can be written as: $a+2b=\frac{13}{9a}$......... (iv)
Solving (i) and (ii), we get
$a=-\frac{1}{3}$ or $-\frac{13}{9}$
If $a=-\frac{13}{9}$, then $b=-\frac{18}{169}$, ( from equation (ii))
But this does not satisfy (i) and (iii).
$\therefore a\ne -\frac{13}{9}$
If $a=-\frac{1}{3}$, then $b=-2$
Hence, roots are $-2,-\frac{1}{3},-\frac{1}{3}$
Option D is correct.