Solve 12/x^2+14=3/x^2+4+x^2

$$\frac { 12 } { x ^ { 2 } } + 14 = \frac { 3 } { x ^ { 2 } } + 4 + x ^ { 2 }$$

Solve for x (complex solution)

$$x=\sqrt{\sqrt{34}+5}\approx 3.291041157$$

$$x=-\sqrt{\sqrt{34}+5}\approx -3.291041157$$

$$x=-i\sqrt{\sqrt{34}-5}\approx -0-0.911565628i$$

$$x=i\sqrt{\sqrt{34}-5}\approx 0.911565628i$$

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}.

$$12+x^{2}\times 14=3+x^{2}\times 4+x^{2}x^{2}$$

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

$$12+x^{2}\times 14=3+x^{2}\times 4+x^{4}$$

Subtract 3 from both sides.

$$12+x^{2}\times 14-3=x^{2}\times 4+x^{4}$$

Subtract 3 from 12 to get 9.

$$9+x^{2}\times 14=x^{2}\times 4+x^{4}$$

Subtract x^{2}\times 4 from both sides.

$$9+x^{2}\times 14-x^{2}\times 4=x^{4}$$

Combine x^{2}\times 14 and -x^{2}\times 4 to get 10x^{2}.

$$9+10x^{2}=x^{4}$$

Subtract x^{4} from both sides.

$$9+10x^{2}-x^{4}=0$$

Substitute t for x^{2}.

$$-t^{2}+10t+9=0$$

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 -1 for a, 10 for b, and 9 for c in the quadratic formula.

$$t=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\times 9}}{-2}$$

Do the calculations.

$$t=\frac{-10±2\sqrt{34}}{-2}$$

Solve the equation t=\frac{-10±2\sqrt{34}}{-2} when ± is plus and when ± is minus.

$$t=5-\sqrt{34} t=\sqrt{34}+5$$

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

$$x=-i\sqrt{-\left(5-\sqrt{34}\right)} x=i\sqrt{-\left(5-\sqrt{34}\right)} x=-\sqrt{\sqrt{34}+5} x=\sqrt{\sqrt{34}+5}$$

Solve for x

$$x=\sqrt{\sqrt{34}+5}\approx 3.291041157$$

$$x=-\sqrt{\sqrt{34}+5}\approx -3.291041157$$

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}.

$$12+x^{2}\times 14=3+x^{2}\times 4+x^{2}x^{2}$$

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

$$12+x^{2}\times 14=3+x^{2}\times 4+x^{4}$$

Subtract 3 from both sides.

$$12+x^{2}\times 14-3=x^{2}\times 4+x^{4}$$

Subtract 3 from 12 to get 9.

$$9+x^{2}\times 14=x^{2}\times 4+x^{4}$$

Subtract x^{2}\times 4 from both sides.

$$9+x^{2}\times 14-x^{2}\times 4=x^{4}$$

Combine x^{2}\times 14 and -x^{2}\times 4 to get 10x^{2}.

$$9+10x^{2}=x^{4}$$

Subtract x^{4} from both sides.

$$9+10x^{2}-x^{4}=0$$

Substitute t for x^{2}.

$$-t^{2}+10t+9=0$$

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 -1 for a, 10 for b, and 9 for c in the quadratic formula.

$$t=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\times 9}}{-2}$$

Do the calculations.

$$t=\frac{-10±2\sqrt{34}}{-2}$$

Solve the equation t=\frac{-10±2\sqrt{34}}{-2} when ± is plus and when ± is minus.

$$t=5-\sqrt{34} t=\sqrt{34}+5$$

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

$$x=\sqrt{\sqrt{34}+5} x=-\sqrt{\sqrt{34}+5}$$