Solve (x+2)sqrt{x-6}=0

$$(x+2) \sqrt{ x-6 } =0$$

Solve for x (complex solution)

$$x=6$$

$$x=-2$$

Solution Steps

Use the distributive property to multiply x+2 by \sqrt{x-6}.

$$x\sqrt{x-6}+2\sqrt{x-6}=0$$

Subtract 2\sqrt{x-6} from both sides of the equation.

$$x\sqrt{x-6}=-2\sqrt{x-6}$$

Square both sides of the equation.

$$\left(x\sqrt{x-6}\right)^{2}=\left(-2\sqrt{x-6}\right)^{2}$$

Expand \left(x\sqrt{x-6}\right)^{2}.

$$x^{2}\left(\sqrt{x-6}\right)^{2}=\left(-2\sqrt{x-6}\right)^{2}$$

Calculate \sqrt{x-6} to the power of 2 and get x-6.

$$x^{2}\left(x-6\right)=\left(-2\sqrt{x-6}\right)^{2}$$

Use the distributive property to multiply x^{2} by x-6.

$$x^{3}-6x^{2}=\left(-2\sqrt{x-6}\right)^{2}$$

Expand \left(-2\sqrt{x-6}\right)^{2}.

$$x^{3}-6x^{2}=\left(-2\right)^{2}\left(\sqrt{x-6}\right)^{2}$$

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

$$x^{3}-6x^{2}=4\left(\sqrt{x-6}\right)^{2}$$

Calculate \sqrt{x-6} to the power of 2 and get x-6.

$$x^{3}-6x^{2}=4\left(x-6\right)$$

Use the distributive property to multiply 4 by x-6.

$$x^{3}-6x^{2}=4x-24$$

Subtract 4x from both sides.

$$x^{3}-6x^{2}-4x=-24$$

Add 24 to both sides.

$$x^{3}-6x^{2}-4x+24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-6x^{2}-4x+24 by x-2 to get x^{2}-4x-12. Solve the equation where the result equals to 0.

$$x^{2}-4x-12=0$$

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, -4 for b, and -12 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$x=\frac{4±8}{2}$$

Solve the equation x^{2}-4x-12=0 when ± is plus and when ± is minus.

$$x=-2 x=6$$

List all found solutions.

$$x=2 x=-2 x=6$$

Substitute 2 for x in the equation \left(x+2\right)\sqrt{x-6}=0.

$$\left(2+2\right)\sqrt{2-6}=0$$

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

$$8i=0$$

Substitute -2 for x in the equation \left(x+2\right)\sqrt{x-6}=0.

$$\left(-2+2\right)\sqrt{-2-6}=0$$

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

$$0=0$$

Substitute 6 for x in the equation \left(x+2\right)\sqrt{x-6}=0.

$$\left(6+2\right)\sqrt{6-6}=0$$

Simplify. The value x=6 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{x-6}x=-2\sqrt{x-6}.

$$x=-2 x=6$$

Solve for x

$$x=6$$

Solution Steps

Use the distributive property to multiply x+2 by \sqrt{x-6}.

$$x\sqrt{x-6}+2\sqrt{x-6}=0$$

Subtract 2\sqrt{x-6} from both sides of the equation.

$$x\sqrt{x-6}=-2\sqrt{x-6}$$

Square both sides of the equation.

$$\left(x\sqrt{x-6}\right)^{2}=\left(-2\sqrt{x-6}\right)^{2}$$

Expand \left(x\sqrt{x-6}\right)^{2}.

$$x^{2}\left(\sqrt{x-6}\right)^{2}=\left(-2\sqrt{x-6}\right)^{2}$$

Calculate \sqrt{x-6} to the power of 2 and get x-6.

$$x^{2}\left(x-6\right)=\left(-2\sqrt{x-6}\right)^{2}$$

Use the distributive property to multiply x^{2} by x-6.

$$x^{3}-6x^{2}=\left(-2\sqrt{x-6}\right)^{2}$$

Expand \left(-2\sqrt{x-6}\right)^{2}.

$$x^{3}-6x^{2}=\left(-2\right)^{2}\left(\sqrt{x-6}\right)^{2}$$

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

$$x^{3}-6x^{2}=4\left(\sqrt{x-6}\right)^{2}$$

Calculate \sqrt{x-6} to the power of 2 and get x-6.

$$x^{3}-6x^{2}=4\left(x-6\right)$$

Use the distributive property to multiply 4 by x-6.

$$x^{3}-6x^{2}=4x-24$$

Subtract 4x from both sides.

$$x^{3}-6x^{2}-4x=-24$$

Add 24 to both sides.

$$x^{3}-6x^{2}-4x+24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-6x^{2}-4x+24 by x-2 to get x^{2}-4x-12. Solve the equation where the result equals to 0.

$$x^{2}-4x-12=0$$

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, -4 for b, and -12 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$x=\frac{4±8}{2}$$

Solve the equation x^{2}-4x-12=0 when ± is plus and when ± is minus.

$$x=-2 x=6$$

List all found solutions.

$$x=2 x=-2 x=6$$

Substitute 2 for x in the equation \left(x+2\right)\sqrt{x-6}=0. The expression \sqrt{2-6} is undefined because the radicand cannot be negative.

$$\left(2+2\right)\sqrt{2-6}=0$$

Substitute -2 for x in the equation \left(x+2\right)\sqrt{x-6}=0. The expression \sqrt{-2-6} is undefined because the radicand cannot be negative.

$$\left(-2+2\right)\sqrt{-2-6}=0$$

Substitute 6 for x in the equation \left(x+2\right)\sqrt{x-6}=0.

$$\left(6+2\right)\sqrt{6-6}=0$$

Simplify. The value x=6 satisfies the equation.

$$0=0$$

Equation \sqrt{x-6}x=-2\sqrt{x-6} has a unique solution.

$$x=6$$