Solve sqrt{x-3}*x-2/x^2+9=0

$$\sqrt{ x-3 } \times \frac{ x-2 }{ { x }^{ 2 } +9 } =0$$

Solve for x (complex solution)

$$x=3$$

$$x=2$$

Solution Steps

Variable x cannot be equal to any of the values -3i,3i since division by zero is not defined. Multiply both sides of the equation by \left(x-3i\right)\left(x+3i\right).

$$\sqrt{x-3}\left(x-2\right)=0$$

Use the distributive property to multiply \sqrt{x-3} by x-2.

$$\sqrt{x-3}x-2\sqrt{x-3}=0$$

Subtract -2\sqrt{x-3} from both sides of the equation.

$$\sqrt{x-3}x=2\sqrt{x-3}$$

Square both sides of the equation.

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

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

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

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

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

Use the distributive property to multiply x-3 by x^{2}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by x-3.

$$x^{3}-3x^{2}=4x-12$$

Subtract 4x from both sides.

$$x^{3}-3x^{2}-4x=-12$$

Add 12 to both sides.

$$x^{3}-3x^{2}-4x+12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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}-3x^{2}-4x+12 by x-2 to get x^{2}-x-6. Solve the equation where the result equals to 0.

$$x^{2}-x-6=0$$

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, -1 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-6\right)}}{2}$$

Do the calculations.

$$x=\frac{1±5}{2}$$

Solve the equation x^{2}-x-6=0 when ± is plus and when ± is minus.

$$x=-2 x=3$$

List all found solutions.

$$x=2 x=-2 x=3$$

Substitute 2 for x in the equation \sqrt{x-3}\times \frac{x-2}{x^{2}+9}=0.

$$\sqrt{2-3}\times \frac{2-2}{2^{2}+9}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

Substitute -2 for x in the equation \sqrt{x-3}\times \frac{x-2}{x^{2}+9}=0.

$$\sqrt{-2-3}\times \frac{-2-2}{\left(-2\right)^{2}+9}=0$$

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

$$-\frac{4}{13}i\times 5^{\frac{1}{2}}=0$$

Substitute 3 for x in the equation \sqrt{x-3}\times \frac{x-2}{x^{2}+9}=0.

$$\sqrt{3-3}\times \frac{3-2}{3^{2}+9}=0$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{x-3}x=2\sqrt{x-3}.

$$x=2 x=3$$

Solve for x

$$x=3$$

Solution Steps

Multiply both sides of the equation by x^{2}+9.

$$\sqrt{x-3}\left(x-2\right)=0$$

Use the distributive property to multiply \sqrt{x-3} by x-2.

$$\sqrt{x-3}x-2\sqrt{x-3}=0$$

Subtract -2\sqrt{x-3} from both sides of the equation.

$$\sqrt{x-3}x=2\sqrt{x-3}$$

Square both sides of the equation.

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

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

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

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

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

Use the distributive property to multiply x-3 by x^{2}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by x-3.

$$x^{3}-3x^{2}=4x-12$$

Subtract 4x from both sides.

$$x^{3}-3x^{2}-4x=-12$$

Add 12 to both sides.

$$x^{3}-3x^{2}-4x+12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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}-3x^{2}-4x+12 by x-2 to get x^{2}-x-6. Solve the equation where the result equals to 0.

$$x^{2}-x-6=0$$

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, -1 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-6\right)}}{2}$$

Do the calculations.

$$x=\frac{1±5}{2}$$

Solve the equation x^{2}-x-6=0 when ± is plus and when ± is minus.

$$x=-2 x=3$$

List all found solutions.

$$x=2 x=-2 x=3$$

Substitute 2 for x in the equation \sqrt{x-3}\times \frac{x-2}{x^{2}+9}=0. The expression \sqrt{2-3} is undefined because the radicand cannot be negative.

$$\sqrt{2-3}\times \frac{2-2}{2^{2}+9}=0$$

Substitute -2 for x in the equation \sqrt{x-3}\times \frac{x-2}{x^{2}+9}=0. The expression \sqrt{-2-3} is undefined because the radicand cannot be negative.

$$\sqrt{-2-3}\times \frac{-2-2}{\left(-2\right)^{2}+9}=0$$

Substitute 3 for x in the equation \sqrt{x-3}\times \frac{x-2}{x^{2}+9}=0.

$$\sqrt{3-3}\times \frac{3-2}{3^{2}+9}=0$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

Equation \sqrt{x-3}x=2\sqrt{x-3} has a unique solution.

$$x=3$$