Solve (x+3)sqrt{x-1}=0

$$( x + 3 ) \sqrt { x - 1 } = 0$$

Solve for x (complex solution)

$$x=1$$

$$x=-3$$

Solution Steps

Use the distributive property to multiply x+3 by \sqrt{x-1}.

$$x\sqrt{x-1}+3\sqrt{x-1}=0$$

Subtract 3\sqrt{x-1} from both sides of the equation.

$$x\sqrt{x-1}=-3\sqrt{x-1}$$

Square both sides of the equation.

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

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

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

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

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

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

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

Expand \left(-3\sqrt{x-1}\right)^{2}.

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

Calculate -3 to the power of 2 and get 9.

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

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

$$x^{3}-x^{2}=9\left(x-1\right)$$

Use the distributive property to multiply 9 by x-1.

$$x^{3}-x^{2}=9x-9$$

Subtract 9x from both sides.

$$x^{3}-x^{2}-9x=-9$$

Add 9 to both sides.

$$x^{3}-x^{2}-9x+9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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}-9x+9 by x-1 to get x^{2}-9. Solve the equation where the result equals to 0.

$$x^{2}-9=0$$

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 -9 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$x=\frac{0±6}{2}$$

Solve the equation x^{2}-9=0 when ± is plus and when ± is minus.

$$x=-3 x=3$$

List all found solutions.

$$x=1 x=-3 x=3$$

Substitute 1 for x in the equation \left(x+3\right)\sqrt{x-1}=0.

$$\left(1+3\right)\sqrt{1-1}=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute -3 for x in the equation \left(x+3\right)\sqrt{x-1}=0.

$$\left(-3+3\right)\sqrt{-3-1}=0$$

Simplify. The value x=-3 satisfies the equation.

$$0=0$$

Substitute 3 for x in the equation \left(x+3\right)\sqrt{x-1}=0.

$$\left(3+3\right)\sqrt{3-1}=0$$

Simplify. The value x=3 does not satisfy the equation.

$$6\times 2^{\frac{1}{2}}=0$$

List all solutions of \sqrt{x-1}x=-3\sqrt{x-1}.

$$x=1 x=-3$$

Solve for x

$$x=1$$

Solution Steps

Use the distributive property to multiply x+3 by \sqrt{x-1}.

$$x\sqrt{x-1}+3\sqrt{x-1}=0$$

Subtract 3\sqrt{x-1} from both sides of the equation.

$$x\sqrt{x-1}=-3\sqrt{x-1}$$

Square both sides of the equation.

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

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

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

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

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

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

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

Expand \left(-3\sqrt{x-1}\right)^{2}.

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

Calculate -3 to the power of 2 and get 9.

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

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

$$x^{3}-x^{2}=9\left(x-1\right)$$

Use the distributive property to multiply 9 by x-1.

$$x^{3}-x^{2}=9x-9$$

Subtract 9x from both sides.

$$x^{3}-x^{2}-9x=-9$$

Add 9 to both sides.

$$x^{3}-x^{2}-9x+9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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}-9x+9 by x-1 to get x^{2}-9. Solve the equation where the result equals to 0.

$$x^{2}-9=0$$

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 -9 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$x=\frac{0±6}{2}$$

Solve the equation x^{2}-9=0 when ± is plus and when ± is minus.

$$x=-3 x=3$$

List all found solutions.

$$x=1 x=-3 x=3$$

Substitute 1 for x in the equation \left(x+3\right)\sqrt{x-1}=0.

$$\left(1+3\right)\sqrt{1-1}=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute -3 for x in the equation \left(x+3\right)\sqrt{x-1}=0. The expression \sqrt{-3-1} is undefined because the radicand cannot be negative.

$$\left(-3+3\right)\sqrt{-3-1}=0$$

Substitute 3 for x in the equation \left(x+3\right)\sqrt{x-1}=0.

$$\left(3+3\right)\sqrt{3-1}=0$$

Simplify. The value x=3 does not satisfy the equation.

$$6\times 2^{\frac{1}{2}}=0$$

Equation \sqrt{x-1}x=-3\sqrt{x-1} has a unique solution.

$$x=1$$