Solve 2y^2-8/y^2=6-2y^2

$$\frac { 2 y ^ { 2 } - 8 } { y ^ { 2 } } = 6 - 2 y ^ { 2 }$$

Solve for y (complex solution)

$$y=\sqrt{\sqrt{5}+1}\approx 1.79890744$$

$$y=-\sqrt{\sqrt{5}+1}\approx -1.79890744$$

$$y=-i\sqrt{\sqrt{5}-1}\approx -0-1.111785941i$$

$$y=i\sqrt{\sqrt{5}-1}\approx 1.111785941i$$

Steps Using the Quadratic Formula

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

$$2y^{2}-8=y^{2}\times 6-2y^{2}y^{2}$$

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

$$2y^{2}-8=y^{2}\times 6-2y^{4}$$

Subtract y^{2}\times 6 from both sides.

$$2y^{2}-8-y^{2}\times 6=-2y^{4}$$

Combine 2y^{2} and -y^{2}\times 6 to get -4y^{2}.

$$-4y^{2}-8=-2y^{4}$$

Add 2y^{4} to both sides.

$$-4y^{2}-8+2y^{4}=0$$

Substitute t for y^{2}.

$$2t^{2}-4t-8=0$$

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 2 for a, -4 for b, and -8 for c in the quadratic formula.

$$t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Do the calculations.

$$t=\frac{4±4\sqrt{5}}{4}$$

Solve the equation t=\frac{4±4\sqrt{5}}{4} when ± is plus and when ± is minus.

$$t=\sqrt{5}+1 t=1-\sqrt{5}$$

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

$$y=-\sqrt{\sqrt{5}+1} y=\sqrt{\sqrt{5}+1} y=-i\sqrt{-\left(1-\sqrt{5}\right)} y=i\sqrt{-\left(1-\sqrt{5}\right)}$$

Solve for y

$$y=\sqrt{\sqrt{5}+1}\approx 1.79890744$$

$$y=-\sqrt{\sqrt{5}+1}\approx -1.79890744$$

Steps Using the Quadratic Formula

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

$$2y^{2}-8=y^{2}\times 6-2y^{2}y^{2}$$

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

$$2y^{2}-8=y^{2}\times 6-2y^{4}$$

Subtract y^{2}\times 6 from both sides.

$$2y^{2}-8-y^{2}\times 6=-2y^{4}$$

Combine 2y^{2} and -y^{2}\times 6 to get -4y^{2}.

$$-4y^{2}-8=-2y^{4}$$

Add 2y^{4} to both sides.

$$-4y^{2}-8+2y^{4}=0$$

Substitute t for y^{2}.

$$2t^{2}-4t-8=0$$

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 2 for a, -4 for b, and -8 for c in the quadratic formula.

$$t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Do the calculations.

$$t=\frac{4±4\sqrt{5}}{4}$$

Solve the equation t=\frac{4±4\sqrt{5}}{4} when ± is plus and when ± is minus.

$$t=\sqrt{5}+1 t=1-\sqrt{5}$$

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

$$y=\sqrt{\sqrt{5}+1} y=-\sqrt{\sqrt{5}+1}$$