Solve 6x^2-5-4/x^2=0

$$6 x ^ { 2 } - 5 - \frac { 4 } { x ^ { 2 } } = 0$$

Solve for x (complex solution)

$$x=\frac{\sqrt{2}i}{2}\approx 0.707106781i$$

$$x=-\frac{\sqrt{2}i}{2}\approx -0-0.707106781i$$

$$x = -\frac{2 \sqrt{3}}{3} \approx -1.154700538$$

$$x = \frac{2 \sqrt{3}}{3} \approx 1.154700538$$

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}.

$$6x^{2}x^{2}+x^{2}\left(-5\right)-4=0$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$6x^{4}+x^{2}\left(-5\right)-4=0$$

Substitute t for x^{2}.

$$6t^{2}-5t-4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 6 for a, -5 for b, and -4 for c in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

Do the calculations.

$$t=\frac{5±11}{12}$$

Solve the equation t=\frac{5±11}{12} when ± is plus and when ± is minus.

$$t=\frac{4}{3} t=-\frac{1}{2}$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for each t.

$$x=-\frac{2\sqrt{3}}{3} x=\frac{2\sqrt{3}}{3} x=-\frac{\sqrt{2}i}{2} x=\frac{\sqrt{2}i}{2}$$

Solve for x

$$x = \frac{2 \sqrt{3}}{3} \approx 1.154700538$$

$$x = -\frac{2 \sqrt{3}}{3} \approx -1.154700538$$

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}.

$$6x^{2}x^{2}+x^{2}\left(-5\right)-4=0$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$6x^{4}+x^{2}\left(-5\right)-4=0$$

Substitute t for x^{2}.

$$6t^{2}-5t-4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 6 for a, -5 for b, and -4 for c in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

Do the calculations.

$$t=\frac{5±11}{12}$$

Solve the equation t=\frac{5±11}{12} when ± is plus and when ± is minus.

$$t=\frac{4}{3} t=-\frac{1}{2}$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x=\frac{2\sqrt{3}}{3} x=-\frac{2\sqrt{3}}{3}$$