Solve sqrt{2x^3+2}=x^+1

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

Solve for x

$$x=-1$$

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

$$x=1$$

Solution Steps

Square both sides of the equation.

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

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

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

Calculate x to the power of 1 and get x.

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

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

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

Subtract 1 from both sides.

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

Subtract 1 from 2 to get 1.

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

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

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

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 2. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-x^{2}-2x+1 by x-1 to get 2x^{2}+x-1. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{-1±3}{4}$$

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

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

List all found solutions.

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

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

$$\sqrt{2\times 1^{3}+2}=1^{1}+1$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

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

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

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

$$0=0$$

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

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

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

$$\frac{3}{2}=\frac{3}{2}$$

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

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