Solve sqrt{1+x^2}*sqrt{{left(4x/x^2+1right)}^2-8/x^2+1}=sqrt{x^2-1}

$$\sqrt{ 1+ { x }^{ 2 } } \times \sqrt{ { \left( \frac{ 4x }{ { x }^{ 2 } +1 } \right) }^{ 2 } - \frac{ 8 }{ { x }^{ 2 } +1 } } = \sqrt{ { x }^{ 2 } -1 }$$

Solve for x

$$x=1$$

$$x=\sqrt{7}\approx 2.645751311$$

$$x=-\sqrt{7}\approx -2.645751311$$

$$x=-1$$

Solution Steps

Square both sides of the equation.

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

To raise \frac{4x}{x^{2}+1} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x^{2}+1\right)^{2} and x^{2}+1 is \left(x^{2}+1\right)^{2}. Multiply \frac{8}{x^{2}+1} times \frac{x^{2}+1}{x^{2}+1}.

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

Since \frac{\left(4x\right)^{2}}{\left(x^{2}+1\right)^{2}} and \frac{8\left(x^{2}+1\right)}{\left(x^{2}+1\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(4x\right)^{2}-8\left(x^{2}+1\right).

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

Combine like terms in \left(4x\right)^{2}-8x^{2}-8.

$$\left(\sqrt{1+x^{2}}\sqrt{\frac{8x^{2}-8}{\left(x^{2}+1\right)^{2}}}\right)^{2}=\left(\sqrt{x^{2}-1}\right)^{2}$$

Expand \left(\sqrt{1+x^{2}}\sqrt{\frac{8x^{2}-8}{\left(x^{2}+1\right)^{2}}}\right)^{2}.

$$\left(\sqrt{1+x^{2}}\right)^{2}\left(\sqrt{\frac{8x^{2}-8}{\left(x^{2}+1\right)^{2}}}\right)^{2}=\left(\sqrt{x^{2}-1}\right)^{2}$$

Calculate \sqrt{1+x^{2}} to the power of 2 and get 1+x^{2}.

$$\left(1+x^{2}\right)\left(\sqrt{\frac{8x^{2}-8}{\left(x^{2}+1\right)^{2}}}\right)^{2}=\left(\sqrt{x^{2}-1}\right)^{2}$$

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

$$\left(1+x^{2}\right)\left(\sqrt{\frac{8x^{2}-8}{\left(x^{2}\right)^{2}+2x^{2}+1}}\right)^{2}=\left(\sqrt{x^{2}-1}\right)^{2}$$

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

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

Calculate \sqrt{\frac{8x^{2}-8}{x^{4}+2x^{2}+1}} to the power of 2 and get \frac{8x^{2}-8}{x^{4}+2x^{2}+1}.

$$\left(1+x^{2}\right)\times \frac{8x^{2}-8}{x^{4}+2x^{2}+1}=\left(\sqrt{x^{2}-1}\right)^{2}$$

Express \left(1+x^{2}\right)\times \frac{8x^{2}-8}{x^{4}+2x^{2}+1} as a single fraction.

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

Calculate \sqrt{x^{2}-1} to the power of 2 and get x^{2}-1.

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

Factor the expressions that are not already factored in \frac{\left(1+x^{2}\right)\left(8x^{2}-8\right)}{x^{4}+2x^{2}+1}.

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

Cancel out x^{2}+1 in both numerator and denominator.

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

Multiply both sides of the equation by x^{2}+1.

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

Use the distributive property to multiply 8 by x-1.

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

Use the distributive property to multiply 8x-8 by x+1 and combine like terms.

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

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

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

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

$$8x^{2}-8=x^{4}+x^{2}-x^{2}-1$$

Combine x^{2} and -x^{2} to get 0.

$$8x^{2}-8=x^{4}-1$$

Subtract x^{4} from both sides.

$$8x^{2}-8-x^{4}=-1$$

Add 1 to both sides.

$$8x^{2}-8-x^{4}+1=0$$

Add -8 and 1 to get -7.

$$8x^{2}-7-x^{4}=0$$

Substitute t for x^{2}.

$$-t^{2}+8t-7=0$$

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, 8 for b, and -7 for c in the quadratic formula.

$$t=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-7\right)}}{-2}$$

Do the calculations.

$$t=\frac{-8±6}{-2}$$

Solve the equation t=\frac{-8±6}{-2} when ± is plus and when ± is minus.

$$t=1 t=7$$

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

$$x=1 x=-1 x=\sqrt{7} x=-\sqrt{7}$$

Substitute 1 for x in the equation \sqrt{1+x^{2}}\sqrt{\left(\frac{4x}{x^{2}+1}\right)^{2}-\frac{8}{x^{2}+1}}=\sqrt{x^{2}-1}.

$$\sqrt{1+1^{2}}\sqrt{\left(\frac{4\times 1}{1^{2}+1}\right)^{2}-\frac{8}{1^{2}+1}}=\sqrt{1^{2}-1}$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute -1 for x in the equation \sqrt{1+x^{2}}\sqrt{\left(\frac{4x}{x^{2}+1}\right)^{2}-\frac{8}{x^{2}+1}}=\sqrt{x^{2}-1}.

$$\sqrt{1+\left(-1\right)^{2}}\sqrt{\left(\frac{4\left(-1\right)}{\left(-1\right)^{2}+1}\right)^{2}-\frac{8}{\left(-1\right)^{2}+1}}=\sqrt{\left(-1\right)^{2}-1}$$

Simplify. The value x=-1 satisfies the equation.

$$0=0$$

Substitute \sqrt{7} for x in the equation \sqrt{1+x^{2}}\sqrt{\left(\frac{4x}{x^{2}+1}\right)^{2}-\frac{8}{x^{2}+1}}=\sqrt{x^{2}-1}.

$$\sqrt{1+\left(\sqrt{7}\right)^{2}}\sqrt{\left(\frac{4\sqrt{7}}{\left(\sqrt{7}\right)^{2}+1}\right)^{2}-\frac{8}{\left(\sqrt{7}\right)^{2}+1}}=\sqrt{\left(\sqrt{7}\right)^{2}-1}$$

Simplify. The value x=\sqrt{7} satisfies the equation.

$$6^{\frac{1}{2}}=6^{\frac{1}{2}}$$

Substitute -\sqrt{7} for x in the equation \sqrt{1+x^{2}}\sqrt{\left(\frac{4x}{x^{2}+1}\right)^{2}-\frac{8}{x^{2}+1}}=\sqrt{x^{2}-1}.

$$\sqrt{1+\left(-\sqrt{7}\right)^{2}}\sqrt{\left(\frac{4\left(-\sqrt{7}\right)}{\left(-\sqrt{7}\right)^{2}+1}\right)^{2}-\frac{8}{\left(-\sqrt{7}\right)^{2}+1}}=\sqrt{\left(-\sqrt{7}\right)^{2}-1}$$

Simplify. The value x=-\sqrt{7} satisfies the equation.

$$6^{\frac{1}{2}}=6^{\frac{1}{2}}$$

List all solutions of \sqrt{\left(\frac{4x}{x^{2}+1}\right)^{2}-\frac{8}{x^{2}+1}}\sqrt{x^{2}+1}=\sqrt{x^{2}-1}.

$$x=1 x=-1 x=\sqrt{7} x=-\sqrt{7}$$