Solve (1+x^2)*(2+x^2)=5

$$( 1 + x ^ { 2 } ) \cdot ( 2 + x ^ { 2 } ) = 5$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{2\left(\sqrt{21}+3\right)}}{2}\approx 1.947122967i$$

$$x=-\frac{i\sqrt{2\left(\sqrt{21}+3\right)}}{2}\approx -0-1.947122967i$$

$$x=-\frac{\sqrt{2\left(\sqrt{21}-3\right)}}{2}\approx -0.889543618$$

$$x=\frac{\sqrt{2\left(\sqrt{21}-3\right)}}{2}\approx 0.889543618$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1+x^{2} by 2+x^{2} and combine like terms.

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

Subtract 5 from both sides.

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

Subtract 5 from 2 to get -3.

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

Substitute t for x^{2}.

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

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 -3 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{-3±\sqrt{21}}{2}$$

Solve the equation t=\frac{-3±\sqrt{21}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{21}-3}{2} t=\frac{-\sqrt{21}-3}{2}$$

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

$$x=-\sqrt{\frac{\sqrt{21}-3}{2}} x=\sqrt{\frac{\sqrt{21}-3}{2}} x=-i\sqrt{\frac{\sqrt{21}+3}{2}} x=i\sqrt{\frac{\sqrt{21}+3}{2}}$$

Solve for x

$$x=-\frac{\sqrt{2\left(\sqrt{21}-3\right)}}{2}\approx -0.889543618$$

$$x=\frac{\sqrt{2\left(\sqrt{21}-3\right)}}{2}\approx 0.889543618$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1+x^{2} by 2+x^{2} and combine like terms.

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

Subtract 5 from both sides.

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

Subtract 5 from 2 to get -3.

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

Substitute t for x^{2}.

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

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 -3 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{-3±\sqrt{21}}{2}$$

Solve the equation t=\frac{-3±\sqrt{21}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{21}-3}{2} t=\frac{-\sqrt{21}-3}{2}$$

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

$$x=\frac{\sqrt{2\sqrt{21}-6}}{2} x=-\frac{\sqrt{2\sqrt{21}-6}}{2}$$