Question

If the equation $(3x{)}^2+(27\times 3^{\frac{1}{k}}-15)x+4=0$ has equal roots, then k =

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

Answer 1

The correct option is B $-\frac{1}{2}$

The given equation will have equal roots if D = 0
$\begin{array}{cc}\Rightarrow & {(27\times 3^{\frac{1}{k}}-15)}^2-144=0\\ \Rightarrow & 27\times 3^{\frac{1}{k}}-15=\pm 12\\ \Rightarrow & 27\times 3^{\frac{1}{k}}=27or3\\ \Rightarrow & 3^{\frac{1}{k}}=1or{3}^{\frac{1}{k}}=\frac{1}{9}\\ \Rightarrow & \frac{1}{k}=0or-2\end{array}$
So $k=-\frac{1}{2}$, inifnity.