Solve {l}{text{If}x^{log_{3}^x^2-(log_3^x)^2}-10=1/x^2text{,then}x=}{{lll}{text{a)}3}&{text{b)}9}&{text{c)}27}}

$$\left. \begin{array} { l } { \text { If } x ^ { \log _ { 3 } ^ { x ^ { 2 } } - ( \log _ { 3 } ^ { x } ) ^ { 2 } } - 10 = \frac { 1 } { x ^ { 2 } } \text { , then } x = } \\ { \left. \begin{array} { l l l } { \text { a) } 3 } & { \text { b) } 9 } & { \text { c) } 27 } \end{array} \right. } \end{array} \right.$$

Solve for x

$$x=-\frac{\sqrt{10}i}{10}\approx -0-0.316227766i$$

$$x=\frac{\sqrt{10}i}{10}\approx 0.316227766i$$

Steps by Finding Square Root

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$-10x^{2}=1$$

Divide both sides by -10.

$$x^{2}=-\frac{1}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{10}i}{10} x=-\frac{\sqrt{10}i}{10}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$-10x^{2}=1$$

Subtract 1 from both sides.

$$-10x^{2}-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 0 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-10\right)\left(-1\right)}}{2\left(-10\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-10\right)\left(-1\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{0±\sqrt{40\left(-1\right)}}{2\left(-10\right)}$$

Multiply 40 times -1.

$$x=\frac{0±\sqrt{-40}}{2\left(-10\right)}$$

Take the square root of -40.

$$x=\frac{0±2\sqrt{10}i}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{0±2\sqrt{10}i}{-20}$$

Now solve the equation x=\frac{0±2\sqrt{10}i}{-20} when ± is plus.

$$x=-\frac{\sqrt{10}i}{10}$$

Now solve the equation x=\frac{0±2\sqrt{10}i}{-20} when ± is minus.

$$x=\frac{\sqrt{10}i}{10}$$

The equation is now solved.

$$x=-\frac{\sqrt{10}i}{10} x=\frac{\sqrt{10}i}{10}$$