Solve x^2(x^2-3)=5

$$x ^ { 2 } ( x ^ { 2 } - 3 ) = 5$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{2\left(\sqrt{29}-3\right)}}{2}\approx 1.092054213i$$

$$x=-\frac{i\sqrt{2\left(\sqrt{29}-3\right)}}{2}\approx -0-1.092054213i$$

$$x = -\frac{\sqrt{2 {(\sqrt{29} + 3)}}}{2} \approx -2.047579645$$

$$x = \frac{\sqrt{2 {(\sqrt{29} + 3)}}}{2} \approx 2.047579645$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x^{2} by x^{2}-3.

$$x^{4}-3x^{2}=5$$

Subtract 5 from both sides.

$$x^{4}-3x^{2}-5=0$$

Substitute t for x^{2}.

$$t^{2}-3t-5=0$$

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, -3 for b, and -5 for c in the quadratic formula.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-5\right)}}{2}$$

Do the calculations.

$$t=\frac{3±\sqrt{29}}{2}$$

Solve the equation t=\frac{3±\sqrt{29}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{29}+3}{2} t=\frac{3-\sqrt{29}}{2}$$

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

$$x=-\sqrt{\frac{\sqrt{29}+3}{2}} x=\sqrt{\frac{\sqrt{29}+3}{2}} x=-i\sqrt{-\frac{3-\sqrt{29}}{2}} x=i\sqrt{-\frac{3-\sqrt{29}}{2}}$$

Solve for x

$$x = -\frac{\sqrt{2 {(\sqrt{29} + 3)}}}{2} \approx -2.047579645$$

$$x = \frac{\sqrt{2 {(\sqrt{29} + 3)}}}{2} \approx 2.047579645$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x^{2} by x^{2}-3.

$$x^{4}-3x^{2}=5$$

Subtract 5 from both sides.

$$x^{4}-3x^{2}-5=0$$

Substitute t for x^{2}.

$$t^{2}-3t-5=0$$

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, -3 for b, and -5 for c in the quadratic formula.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-5\right)}}{2}$$

Do the calculations.

$$t=\frac{3±\sqrt{29}}{2}$$

Solve the equation t=\frac{3±\sqrt{29}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{29}+3}{2} t=\frac{3-\sqrt{29}}{2}$$

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

$$x=\frac{\sqrt{2\sqrt{29}+6}}{2} x=-\frac{\sqrt{2\sqrt{29}+6}}{2}$$