Solve (x+5)sqrt{x-2}=0

$$( x + 5 ) \sqrt { x - 2 } = 0$$

Solve for x (complex solution)

$$x=2$$

$$x=-5$$

Solution Steps

Use the distributive property to multiply x+5 by \sqrt{x-2}.

$$x\sqrt{x-2}+5\sqrt{x-2}=0$$

Subtract 5\sqrt{x-2} from both sides of the equation.

$$x\sqrt{x-2}=-5\sqrt{x-2}$$

Square both sides of the equation.

$$\left(x\sqrt{x-2}\right)^{2}=\left(-5\sqrt{x-2}\right)^{2}$$

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

$$x^{2}\left(\sqrt{x-2}\right)^{2}=\left(-5\sqrt{x-2}\right)^{2}$$

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

$$x^{2}\left(x-2\right)=\left(-5\sqrt{x-2}\right)^{2}$$

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

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

Expand \left(-5\sqrt{x-2}\right)^{2}.

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

Calculate -5 to the power of 2 and get 25.

$$x^{3}-2x^{2}=25\left(\sqrt{x-2}\right)^{2}$$

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

$$x^{3}-2x^{2}=25\left(x-2\right)$$

Use the distributive property to multiply 25 by x-2.

$$x^{3}-2x^{2}=25x-50$$

Subtract 25x from both sides.

$$x^{3}-2x^{2}-25x=-50$$

Add 50 to both sides.

$$x^{3}-2x^{2}-25x+50=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 50 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±50,±25,±10,±5,±2,±1$$

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}-2x^{2}-25x+50 by x-2 to get x^{2}-25. Solve the equation where the result equals to 0.

$$x^{2}-25=0$$

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 -25 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-25\right)}}{2}$$

Do the calculations.

$$x=\frac{0±10}{2}$$

Solve the equation x^{2}-25=0 when ± is plus and when ± is minus.

$$x=-5 x=5$$

List all found solutions.

$$x=2 x=-5 x=5$$

Substitute 2 for x in the equation \left(x+5\right)\sqrt{x-2}=0.

$$\left(2+5\right)\sqrt{2-2}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

Substitute -5 for x in the equation \left(x+5\right)\sqrt{x-2}=0.

$$\left(-5+5\right)\sqrt{-5-2}=0$$

Simplify. The value x=-5 satisfies the equation.

$$0=0$$

Substitute 5 for x in the equation \left(x+5\right)\sqrt{x-2}=0.

$$\left(5+5\right)\sqrt{5-2}=0$$

Simplify. The value x=5 does not satisfy the equation.

$$10\times 3^{\frac{1}{2}}=0$$

List all solutions of \sqrt{x-2}x=-5\sqrt{x-2}.

$$x=2 x=-5$$

Solve for x

$$x=2$$

Solution Steps

Use the distributive property to multiply x+5 by \sqrt{x-2}.

$$x\sqrt{x-2}+5\sqrt{x-2}=0$$

Subtract 5\sqrt{x-2} from both sides of the equation.

$$x\sqrt{x-2}=-5\sqrt{x-2}$$

Square both sides of the equation.

$$\left(x\sqrt{x-2}\right)^{2}=\left(-5\sqrt{x-2}\right)^{2}$$

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

$$x^{2}\left(\sqrt{x-2}\right)^{2}=\left(-5\sqrt{x-2}\right)^{2}$$

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

$$x^{2}\left(x-2\right)=\left(-5\sqrt{x-2}\right)^{2}$$

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

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

Expand \left(-5\sqrt{x-2}\right)^{2}.

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

Calculate -5 to the power of 2 and get 25.

$$x^{3}-2x^{2}=25\left(\sqrt{x-2}\right)^{2}$$

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

$$x^{3}-2x^{2}=25\left(x-2\right)$$

Use the distributive property to multiply 25 by x-2.

$$x^{3}-2x^{2}=25x-50$$

Subtract 25x from both sides.

$$x^{3}-2x^{2}-25x=-50$$

Add 50 to both sides.

$$x^{3}-2x^{2}-25x+50=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 50 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±50,±25,±10,±5,±2,±1$$

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}-2x^{2}-25x+50 by x-2 to get x^{2}-25. Solve the equation where the result equals to 0.

$$x^{2}-25=0$$

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 -25 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-25\right)}}{2}$$

Do the calculations.

$$x=\frac{0±10}{2}$$

Solve the equation x^{2}-25=0 when ± is plus and when ± is minus.

$$x=-5 x=5$$

List all found solutions.

$$x=2 x=-5 x=5$$

Substitute 2 for x in the equation \left(x+5\right)\sqrt{x-2}=0.

$$\left(2+5\right)\sqrt{2-2}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

Substitute -5 for x in the equation \left(x+5\right)\sqrt{x-2}=0. The expression \sqrt{-5-2} is undefined because the radicand cannot be negative.

$$\left(-5+5\right)\sqrt{-5-2}=0$$

Substitute 5 for x in the equation \left(x+5\right)\sqrt{x-2}=0.

$$\left(5+5\right)\sqrt{5-2}=0$$

Simplify. The value x=5 does not satisfy the equation.

$$10\times 3^{\frac{1}{2}}=0$$

Equation \sqrt{x-2}x=-5\sqrt{x-2} has a unique solution.

$$x=2$$