Solve (x^2-16)*sqrt{x-2}=0

$$( x ^ { 2 } - 16 ) \cdot \sqrt { x - 2 } = 0$$

Solve for x

$$x=2$$

$$x=4$$

Solution Steps

Use the distributive property to multiply x^{2}-16 by \sqrt{x-2}.

$$x^{2}\sqrt{x-2}-16\sqrt{x-2}=0$$

Subtract -16\sqrt{x-2} from both sides of the equation.

$$x^{2}\sqrt{x-2}=16\sqrt{x-2}$$

Square both sides of the equation.

$$\left(x^{2}\sqrt{x-2}\right)^{2}=\left(16\sqrt{x-2}\right)^{2}$$

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

$$\left(x^{2}\right)^{2}\left(\sqrt{x-2}\right)^{2}=\left(16\sqrt{x-2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}\left(\sqrt{x-2}\right)^{2}=\left(16\sqrt{x-2}\right)^{2}$$

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

$$x^{4}\left(x-2\right)=\left(16\sqrt{x-2}\right)^{2}$$

Use the distributive property to multiply x^{4} by x-2.

$$x^{5}-2x^{4}=\left(16\sqrt{x-2}\right)^{2}$$

Expand \left(16\sqrt{x-2}\right)^{2}.

$$x^{5}-2x^{4}=16^{2}\left(\sqrt{x-2}\right)^{2}$$

Calculate 16 to the power of 2 and get 256.

$$x^{5}-2x^{4}=256\left(\sqrt{x-2}\right)^{2}$$

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

$$x^{5}-2x^{4}=256\left(x-2\right)$$

Use the distributive property to multiply 256 by x-2.

$$x^{5}-2x^{4}=256x-512$$

Subtract 256x from both sides.

$$x^{5}-2x^{4}-256x=-512$$

Add 512 to both sides.

$$x^{5}-2x^{4}-256x+512=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 512 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±512,±256,±128,±64,±32,±16,±8,±4,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{5}-2x^{4}-256x+512 by x-2 to get x^{4}-256. Solve the equation where the result equals to 0.

$$x^{4}-256=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -256 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±256,±128,±64,±32,±16,±8,±4,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-256 by x-4 to get x^{3}+4x^{2}+16x+64. Solve the equation where the result equals to 0.

$$x^{3}+4x^{2}+16x+64=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 64 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±64,±32,±16,±8,±4,±2,±1$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+4x^{2}+16x+64 by x+4 to get x^{2}+16. Solve the equation where the result equals to 0.

$$x^{2}+16=0$$

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 16 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-64}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=2 x=4 x=-4$$

Substitute 2 for x in the equation \left(x^{2}-16\right)\sqrt{x-2}=0.

$$\left(2^{2}-16\right)\sqrt{2-2}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

Substitute 4 for x in the equation \left(x^{2}-16\right)\sqrt{x-2}=0.

$$\left(4^{2}-16\right)\sqrt{4-2}=0$$

Simplify. The value x=4 satisfies the equation.

$$0=0$$

Substitute -4 for x in the equation \left(x^{2}-16\right)\sqrt{x-2}=0. The expression \sqrt{-4-2} is undefined because the radicand cannot be negative.

$$\left(\left(-4\right)^{2}-16\right)\sqrt{-4-2}=0$$

List all solutions of \sqrt{x-2}x^{2}=16\sqrt{x-2}.

$$x=2 x=4$$

Solve for x (complex solution)

$$x=4$$

$$x=-4$$

$$x=2$$