Solve sqrt{x^4+8x^3+2x^2-1}=sqrt{x^4+2{x}^2}

$$\sqrt{ { x }^{ 4 } +8 { x }^{ 3 } +2 { x }^{ 2 } -1 } = \sqrt{ { x }^{ 4 } +2 { x }^{ 2 } }$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{3}i-1}{4}\approx -0.25-0.433012702i$$

$$x=\frac{1}{2}=0.5$$

$$x=\frac{-1+\sqrt{3}i}{4}\approx -0.25+0.433012702i$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{4}+8x^{3}+2x^{2}-1}\right)^{2}=\left(\sqrt{x^{4}+2x^{2}}\right)^{2}$$

Calculate \sqrt{x^{4}+8x^{3}+2x^{2}-1} to the power of 2 and get x^{4}+8x^{3}+2x^{2}-1.

$$x^{4}+8x^{3}+2x^{2}-1=\left(\sqrt{x^{4}+2x^{2}}\right)^{2}$$

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

$$x^{4}+8x^{3}+2x^{2}-1=x^{4}+2x^{2}$$

Subtract x^{4} from both sides.

$$x^{4}+8x^{3}+2x^{2}-1-x^{4}=2x^{2}$$

Combine x^{4} and -x^{4} to get 0.

$$8x^{3}+2x^{2}-1=2x^{2}$$

Subtract 2x^{2} from both sides.

$$8x^{3}+2x^{2}-1-2x^{2}=0$$

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

$$8x^{3}-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=\frac{1}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}-1 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 4x^{2}+2x+1. Solve the equation where the result equals to 0.

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

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 4 for a, 2 for b, and 1 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 4\times 1}}{2\times 4}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-12}}{8}$$

Solve the equation 4x^{2}+2x+1=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{3}i-1}{4} x=\frac{-1+\sqrt{3}i}{4}$$

List all found solutions.

$$x=\frac{1}{2} x=\frac{-\sqrt{3}i-1}{4} x=\frac{-1+\sqrt{3}i}{4}$$

Substitute \frac{1}{2} for x in the equation \sqrt{x^{4}+8x^{3}+2x^{2}-1}=\sqrt{x^{4}+2x^{2}}.

$$\sqrt{\left(\frac{1}{2}\right)^{4}+8\times \left(\frac{1}{2}\right)^{3}+2\times \left(\frac{1}{2}\right)^{2}-1}=\sqrt{\left(\frac{1}{2}\right)^{4}+2\times \left(\frac{1}{2}\right)^{2}}$$

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$\frac{3}{4}=\frac{3}{4}$$

Substitute \frac{-\sqrt{3}i-1}{4} for x in the equation \sqrt{x^{4}+8x^{3}+2x^{2}-1}=\sqrt{x^{4}+2x^{2}}.

$$\sqrt{\left(\frac{-\sqrt{3}i-1}{4}\right)^{4}+8\times \left(\frac{-\sqrt{3}i-1}{4}\right)^{3}+2\times \left(\frac{-\sqrt{3}i-1}{4}\right)^{2}-1}=\sqrt{\left(\frac{-\sqrt{3}i-1}{4}\right)^{4}+2\times \left(\frac{-\sqrt{3}i-1}{4}\right)^{2}}$$

Simplify. The value x=\frac{-\sqrt{3}i-1}{4} satisfies the equation.

$$\frac{1}{16}\left(-72+56i\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}=\frac{1}{16}\left(-72+56i\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}$$

Substitute \frac{-1+\sqrt{3}i}{4} for x in the equation \sqrt{x^{4}+8x^{3}+2x^{2}-1}=\sqrt{x^{4}+2x^{2}}.

$$\sqrt{\left(\frac{-1+\sqrt{3}i}{4}\right)^{4}+8\times \left(\frac{-1+\sqrt{3}i}{4}\right)^{3}+2\times \left(\frac{-1+\sqrt{3}i}{4}\right)^{2}-1}=\sqrt{\left(\frac{-1+\sqrt{3}i}{4}\right)^{4}+2\times \left(\frac{-1+\sqrt{3}i}{4}\right)^{2}}$$

Simplify. The value x=\frac{-1+\sqrt{3}i}{4} satisfies the equation.

$$\frac{1}{16}\left(-72-56i\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}=\frac{1}{16}\left(-72-56i\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}$$

List all solutions of \sqrt{x^{4}+8x^{3}+2x^{2}-1}=\sqrt{x^{4}+2x^{2}}.

$$x=\frac{1}{2} x=\frac{-\sqrt{3}i-1}{4} x=\frac{-1+\sqrt{3}i}{4}$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{4}+8x^{3}+2x^{2}-1}\right)^{2}=\left(\sqrt{x^{4}+2x^{2}}\right)^{2}$$

Calculate \sqrt{x^{4}+8x^{3}+2x^{2}-1} to the power of 2 and get x^{4}+8x^{3}+2x^{2}-1.

$$x^{4}+8x^{3}+2x^{2}-1=\left(\sqrt{x^{4}+2x^{2}}\right)^{2}$$

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

$$x^{4}+8x^{3}+2x^{2}-1=x^{4}+2x^{2}$$

Subtract x^{4} from both sides.

$$x^{4}+8x^{3}+2x^{2}-1-x^{4}=2x^{2}$$

Combine x^{4} and -x^{4} to get 0.

$$8x^{3}+2x^{2}-1=2x^{2}$$

Subtract 2x^{2} from both sides.

$$8x^{3}+2x^{2}-1-2x^{2}=0$$

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

$$8x^{3}-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=\frac{1}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}-1 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 4x^{2}+2x+1. Solve the equation where the result equals to 0.

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

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 4 for a, 2 for b, and 1 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 4\times 1}}{2\times 4}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-12}}{8}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=\frac{1}{2}$$

Substitute \frac{1}{2} for x in the equation \sqrt{x^{4}+8x^{3}+2x^{2}-1}=\sqrt{x^{4}+2x^{2}}.

$$\sqrt{\left(\frac{1}{2}\right)^{4}+8\times \left(\frac{1}{2}\right)^{3}+2\times \left(\frac{1}{2}\right)^{2}-1}=\sqrt{\left(\frac{1}{2}\right)^{4}+2\times \left(\frac{1}{2}\right)^{2}}$$

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$\frac{3}{4}=\frac{3}{4}$$

Equation \sqrt{x^{4}+8x^{3}+2x^{2}-1}=\sqrt{x^{4}+2x^{2}} has a unique solution.

$$x=\frac{1}{2}$$