Solve 4(x^2+1)(2x^2+1)=5(x^2-1)^2

$$4 ( x ^ { 2 } + 1 ) ( 2 x ^ { 2 } + 1 ) = 5 ( x ^ { 2 } - 1 ) ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{6\sqrt{31}+33}}{3}\approx 2.716341211i$$

$$x=-\frac{i\sqrt{6\sqrt{31}+33}}{3}\approx -0-2.716341211i$$

$$x=-\frac{\sqrt{6\sqrt{31}-33}}{3}\approx -0.212547035$$

$$x=\frac{\sqrt{6\sqrt{31}-33}}{3}\approx 0.212547035$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by x^{2}+1.

$$\left(4x^{2}+4\right)\left(2x^{2}+1\right)=5\left(x^{2}-1\right)^{2}$$

Use the distributive property to multiply 4x^{2}+4 by 2x^{2}+1 and combine like terms.

$$8x^{4}+12x^{2}+4=5\left(x^{2}-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-1\right)^{2}.

$$8x^{4}+12x^{2}+4=5\left(\left(x^{2}\right)^{2}-2x^{2}+1\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$8x^{4}+12x^{2}+4=5\left(x^{4}-2x^{2}+1\right)$$

Use the distributive property to multiply 5 by x^{4}-2x^{2}+1.

$$8x^{4}+12x^{2}+4=5x^{4}-10x^{2}+5$$

Subtract 5x^{4} from both sides.

$$8x^{4}+12x^{2}+4-5x^{4}=-10x^{2}+5$$

Combine 8x^{4} and -5x^{4} to get 3x^{4}.

$$3x^{4}+12x^{2}+4=-10x^{2}+5$$

Add 10x^{2} to both sides.

$$3x^{4}+12x^{2}+4+10x^{2}=5$$

Combine 12x^{2} and 10x^{2} to get 22x^{2}.

$$3x^{4}+22x^{2}+4=5$$

Subtract 5 from both sides.

$$3x^{4}+22x^{2}+4-5=0$$

Subtract 5 from 4 to get -1.

$$3x^{4}+22x^{2}-1=0$$

Substitute t for x^{2}.

$$3t^{2}+22t-1=0$$

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 3 for a, 22 for b, and -1 for c in the quadratic formula.

$$t=\frac{-22±\sqrt{22^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Do the calculations.

$$t=\frac{-22±4\sqrt{31}}{6}$$

Solve the equation t=\frac{-22±4\sqrt{31}}{6} when ± is plus and when ± is minus.

$$t=\frac{2\sqrt{31}-11}{3} t=\frac{-2\sqrt{31}-11}{3}$$

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

$$x=-\sqrt{\frac{2\sqrt{31}-11}{3}} x=\sqrt{\frac{2\sqrt{31}-11}{3}} x=-i\sqrt{\frac{2\sqrt{31}+11}{3}} x=i\sqrt{\frac{2\sqrt{31}+11}{3}}$$

Solve for x

$$x=-\frac{\sqrt{6\sqrt{31}-33}}{3}\approx -0.212547035$$

$$x=\frac{\sqrt{6\sqrt{31}-33}}{3}\approx 0.212547035$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by x^{2}+1.

$$\left(4x^{2}+4\right)\left(2x^{2}+1\right)=5\left(x^{2}-1\right)^{2}$$

Use the distributive property to multiply 4x^{2}+4 by 2x^{2}+1 and combine like terms.

$$8x^{4}+12x^{2}+4=5\left(x^{2}-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-1\right)^{2}.

$$8x^{4}+12x^{2}+4=5\left(\left(x^{2}\right)^{2}-2x^{2}+1\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$8x^{4}+12x^{2}+4=5\left(x^{4}-2x^{2}+1\right)$$

Use the distributive property to multiply 5 by x^{4}-2x^{2}+1.

$$8x^{4}+12x^{2}+4=5x^{4}-10x^{2}+5$$

Subtract 5x^{4} from both sides.

$$8x^{4}+12x^{2}+4-5x^{4}=-10x^{2}+5$$

Combine 8x^{4} and -5x^{4} to get 3x^{4}.

$$3x^{4}+12x^{2}+4=-10x^{2}+5$$

Add 10x^{2} to both sides.

$$3x^{4}+12x^{2}+4+10x^{2}=5$$

Combine 12x^{2} and 10x^{2} to get 22x^{2}.

$$3x^{4}+22x^{2}+4=5$$

Subtract 5 from both sides.

$$3x^{4}+22x^{2}+4-5=0$$

Subtract 5 from 4 to get -1.

$$3x^{4}+22x^{2}-1=0$$

Substitute t for x^{2}.

$$3t^{2}+22t-1=0$$

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 3 for a, 22 for b, and -1 for c in the quadratic formula.

$$t=\frac{-22±\sqrt{22^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Do the calculations.

$$t=\frac{-22±4\sqrt{31}}{6}$$

Solve the equation t=\frac{-22±4\sqrt{31}}{6} when ± is plus and when ± is minus.

$$t=\frac{2\sqrt{31}-11}{3} t=\frac{-2\sqrt{31}-11}{3}$$

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

$$x=\sqrt{\frac{2\sqrt{31}-11}{3}} x=-\sqrt{\frac{2\sqrt{31}-11}{3}}$$