Question

Solve $6\sqrt{x}=\frac{5}{\sqrt{x}}-13$

• A. $-\frac{25}{4},-\frac{1}{9}$
• B. $-\frac{25}{4},\frac{1}{9}$
• C. $\frac{25}{4},-\frac{1}{9}$
• D. $\frac{25}{4},\frac{1}{9}$

Answer 1

The correct option is D $\frac{25}{4},\frac{1}{9}$

Put $\sqrt{x}=y$. Then we have $6y-\frac{5}{y}=-13$ i.e.,

$6y^2+13y-5=0$

i.e., $(2y+5)(3y-1)=0$

i.e., $y=-\frac{5}{2}$ or $\frac{1}{3}$

i.e., $\sqrt{x}=-\frac{5}{2}$ or $\frac{1}{3}$

$\therefore x=\frac{25}{4}$ or $\frac{1}{9}$

the value $x=\frac{1}{9}$ is a root, but $x=\frac{25}{4}$ is not a root, since on substitution it is found that this value of x does not satisfy the given equation. Such roots are called "extraneous roots." They arise because in the course of the solution, we have squared both sides. Care should be taken to see that the roots obtained really satisfy the given equation.