Solve (x+2)*sqrt{x+1}=0

$$(x+2) \times \sqrt{ x+1 } = 0$$

Solve for x (complex solution)

$$x=-1$$

$$x=-2$$

Solution Steps

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

$$x\sqrt{x+1}+2\sqrt{x+1}=0$$

Subtract 2\sqrt{x+1} from both sides of the equation.

$$x\sqrt{x+1}=-2\sqrt{x+1}$$

Square both sides of the equation.

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

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

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

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

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

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

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

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

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

Calculate -2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by x+1.

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

Subtract 4x from both sides.

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

Subtract 4 from both sides.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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 x^{3}+x^{2}-4x-4 by x+1 to get x^{2}-4. Solve the equation where the result equals to 0.

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

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, 0 for b, and -4 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-4\right)}}{2}$$

Do the calculations.

$$x=\frac{0±4}{2}$$

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

$$x=-2 x=2$$

List all found solutions.

$$x=-1 x=-2 x=2$$

Substitute -1 for x in the equation \left(x+2\right)\sqrt{x+1}=0.

$$\left(-1+2\right)\sqrt{-1+1}=0$$

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

$$0=0$$

Substitute -2 for x in the equation \left(x+2\right)\sqrt{x+1}=0.

$$\left(-2+2\right)\sqrt{-2+1}=0$$

Simplify. The value x=-2 satisfies the equation.

$$0=0$$

Substitute 2 for x in the equation \left(x+2\right)\sqrt{x+1}=0.

$$\left(2+2\right)\sqrt{2+1}=0$$

Simplify. The value x=2 does not satisfy the equation.

$$4\times 3^{\frac{1}{2}}=0$$

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

$$x=-1 x=-2$$

Solve for x

$$x=-1$$

Solution Steps

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

$$x\sqrt{x+1}+2\sqrt{x+1}=0$$

Subtract 2\sqrt{x+1} from both sides of the equation.

$$x\sqrt{x+1}=-2\sqrt{x+1}$$

Square both sides of the equation.

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

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

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

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

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

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

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

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

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

Calculate -2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by x+1.

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

Subtract 4x from both sides.

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

Subtract 4 from both sides.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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 x^{3}+x^{2}-4x-4 by x+1 to get x^{2}-4. Solve the equation where the result equals to 0.

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

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, 0 for b, and -4 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-4\right)}}{2}$$

Do the calculations.

$$x=\frac{0±4}{2}$$

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

$$x=-2 x=2$$

List all found solutions.

$$x=-1 x=-2 x=2$$

Substitute -1 for x in the equation \left(x+2\right)\sqrt{x+1}=0.

$$\left(-1+2\right)\sqrt{-1+1}=0$$

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

$$0=0$$

Substitute -2 for x in the equation \left(x+2\right)\sqrt{x+1}=0. The expression \sqrt{-2+1} is undefined because the radicand cannot be negative.

$$\left(-2+2\right)\sqrt{-2+1}=0$$

Substitute 2 for x in the equation \left(x+2\right)\sqrt{x+1}=0.

$$\left(2+2\right)\sqrt{2+1}=0$$

Simplify. The value x=2 does not satisfy the equation.

$$4\times 3^{\frac{1}{2}}=0$$

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

$$x=-1$$