Solve sqrt{left(x+1right)^2+2x^2}=sqrt{left(-3-xright)^2}

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

Solve for x

$$x=\sqrt{5}+1\approx 3.236067977$$

$$x=1-\sqrt{5}\approx -1.236067977$$

Solution Steps

Square both sides of the equation.

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

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

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-3-x\right)^{2}.

$$3x^{2}+2x+1=\left(\sqrt{9+6x+x^{2}}\right)^{2}$$

Calculate \sqrt{9+6x+x^{2}} to the power of 2 and get 9+6x+x^{2}.

$$3x^{2}+2x+1=9+6x+x^{2}$$

Subtract 9 from both sides.

$$3x^{2}+2x+1-9=6x+x^{2}$$

Subtract 9 from 1 to get -8.

$$3x^{2}+2x-8=6x+x^{2}$$

Subtract 6x from both sides.

$$3x^{2}+2x-8-6x=x^{2}$$

Combine 2x and -6x to get -4x.

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

Subtract x^{2} from both sides.

$$3x^{2}-4x-8-x^{2}=0$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-8\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\left(-8\right)}}{2\times 2}$$

Multiply -8 times -8.

$$x=\frac{-\left(-4\right)±\sqrt{16+64}}{2\times 2}$$

Add 16 to 64.

$$x=\frac{-\left(-4\right)±\sqrt{80}}{2\times 2}$$

Take the square root of 80.

$$x=\frac{-\left(-4\right)±4\sqrt{5}}{2\times 2}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{4±4\sqrt{5}}{4}$$

Now solve the equation x=\frac{4±4\sqrt{5}}{4} when ± is plus. Add 4 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}+4}{4}$$

Divide 4+4\sqrt{5} by 4.

$$x=\sqrt{5}+1$$

Now solve the equation x=\frac{4±4\sqrt{5}}{4} when ± is minus. Subtract 4\sqrt{5} from 4.

$$x=\frac{4-4\sqrt{5}}{4}$$

Divide 4-4\sqrt{5} by 4.

$$x=1-\sqrt{5}$$

The equation is now solved.

$$x=\sqrt{5}+1 x=1-\sqrt{5}$$

Substitute \sqrt{5}+1 for x in the equation \sqrt{\left(x+1\right)^{2}+2x^{2}}=\sqrt{\left(-3-x\right)^{2}}.

$$\sqrt{\left(\sqrt{5}+1+1\right)^{2}+2\left(\sqrt{5}+1\right)^{2}}=\sqrt{\left(-3-\left(\sqrt{5}+1\right)\right)^{2}}$$

Simplify. The value x=\sqrt{5}+1 satisfies the equation.

$$4+5^{\frac{1}{2}}=4+5^{\frac{1}{2}}$$

Substitute 1-\sqrt{5} for x in the equation \sqrt{\left(x+1\right)^{2}+2x^{2}}=\sqrt{\left(-3-x\right)^{2}}.

$$\sqrt{\left(1-\sqrt{5}+1\right)^{2}+2\left(1-\sqrt{5}\right)^{2}}=\sqrt{\left(-3-\left(1-\sqrt{5}\right)\right)^{2}}$$

Simplify. The value x=1-\sqrt{5} satisfies the equation.

$$4-5^{\frac{1}{2}}=4-5^{\frac{1}{2}}$$

List all solutions of \sqrt{\left(x+1\right)^{2}+2x^{2}}=\sqrt{\left(-x-3\right)^{2}}.

$$x=\sqrt{5}+1 x=1-\sqrt{5}$$