Solve (x^2-6x-16)*sqrt{x-3}=0

$$( x ^ { 2 } - 6 x - 16 ) \cdot \sqrt { x - 3 } = 0$$

Solve for x

$$x=3$$

$$x=8$$

Solution Steps

Use the distributive property to multiply x^{2}-6x-16 by \sqrt{x-3}.

$$x^{2}\sqrt{x-3}-6x\sqrt{x-3}-16\sqrt{x-3}=0$$

Subtract -6x\sqrt{x-3}-16\sqrt{x-3} from both sides of the equation.

$$x^{2}\sqrt{x-3}=-\left(-6x\sqrt{x-3}-16\sqrt{x-3}\right)$$

To find the opposite of -6x\sqrt{x-3}-16\sqrt{x-3}, find the opposite of each term.

$$x^{2}\sqrt{x-3}=6x\sqrt{x-3}+16\sqrt{x-3}$$

Square both sides of the equation.

$$\left(x^{2}\sqrt{x-3}\right)^{2}=\left(6x\sqrt{x-3}+16\sqrt{x-3}\right)^{2}$$

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

$$\left(x^{2}\right)^{2}\left(\sqrt{x-3}\right)^{2}=\left(6x\sqrt{x-3}+16\sqrt{x-3}\right)^{2}$$

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

$$x^{4}\left(\sqrt{x-3}\right)^{2}=\left(6x\sqrt{x-3}+16\sqrt{x-3}\right)^{2}$$

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

$$x^{4}\left(x-3\right)=\left(6x\sqrt{x-3}+16\sqrt{x-3}\right)^{2}$$

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

$$x^{5}-3x^{4}=\left(6x\sqrt{x-3}+16\sqrt{x-3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6x\sqrt{x-3}+16\sqrt{x-3}\right)^{2}.

$$x^{5}-3x^{4}=36x^{2}\left(\sqrt{x-3}\right)^{2}+192x\sqrt{x-3}\sqrt{x-3}+256\left(\sqrt{x-3}\right)^{2}$$

Multiply \sqrt{x-3} and \sqrt{x-3} to get \left(\sqrt{x-3}\right)^{2}.

$$x^{5}-3x^{4}=36x^{2}\left(\sqrt{x-3}\right)^{2}+192x\left(\sqrt{x-3}\right)^{2}+256\left(\sqrt{x-3}\right)^{2}$$

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

$$x^{5}-3x^{4}=36x^{2}\left(x-3\right)+192x\left(\sqrt{x-3}\right)^{2}+256\left(\sqrt{x-3}\right)^{2}$$

Use the distributive property to multiply 36x^{2} by x-3.

$$x^{5}-3x^{4}=36x^{3}-108x^{2}+192x\left(\sqrt{x-3}\right)^{2}+256\left(\sqrt{x-3}\right)^{2}$$

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

$$x^{5}-3x^{4}=36x^{3}-108x^{2}+192x\left(x-3\right)+256\left(\sqrt{x-3}\right)^{2}$$

Use the distributive property to multiply 192x by x-3.

$$x^{5}-3x^{4}=36x^{3}-108x^{2}+192x^{2}-576x+256\left(\sqrt{x-3}\right)^{2}$$

Combine -108x^{2} and 192x^{2} to get 84x^{2}.

$$x^{5}-3x^{4}=36x^{3}+84x^{2}-576x+256\left(\sqrt{x-3}\right)^{2}$$

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

$$x^{5}-3x^{4}=36x^{3}+84x^{2}-576x+256\left(x-3\right)$$

Use the distributive property to multiply 256 by x-3.

$$x^{5}-3x^{4}=36x^{3}+84x^{2}-576x+256x-768$$

Combine -576x and 256x to get -320x.

$$x^{5}-3x^{4}=36x^{3}+84x^{2}-320x-768$$

Subtract 36x^{3} from both sides.

$$x^{5}-3x^{4}-36x^{3}=84x^{2}-320x-768$$

Subtract 84x^{2} from both sides.

$$x^{5}-3x^{4}-36x^{3}-84x^{2}=-320x-768$$

Add 320x to both sides.

$$x^{5}-3x^{4}-36x^{3}-84x^{2}+320x=-768$$

Add 768 to both sides.

$$x^{5}-3x^{4}-36x^{3}-84x^{2}+320x+768=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 768 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±768,±384,±256,±192,±128,±96,±64,±48,±32,±24,±16,±12,±8,±6,±4,±3,±2,±1$$

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^{5}-3x^{4}-36x^{3}-84x^{2}+320x+768 by x+2 to get x^{4}-5x^{3}-26x^{2}-32x+384. Solve the equation where the result equals to 0.

$$x^{4}-5x^{3}-26x^{2}-32x+384=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 384 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±384,±192,±128,±96,±64,±48,±32,±24,±16,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-5x^{3}-26x^{2}-32x+384 by x-3 to get x^{3}-2x^{2}-32x-128. Solve the equation where the result equals to 0.

$$x^{3}-2x^{2}-32x-128=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -128 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±128,±64,±32,±16,±8,±4,±2,±1$$

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.

$$x=8$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-2x^{2}-32x-128 by x-8 to get x^{2}+6x+16. Solve the equation where the result equals to 0.

$$x^{2}+6x+16=0$$

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, 6 for b, and 16 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$x=\frac{-6±\sqrt{-28}}{2}$$

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.

$$x=-2 x=3 x=8$$

Substitute -2 for x in the equation \left(x^{2}-6x-16\right)\sqrt{x-3}=0. The expression \sqrt{-2-3} is undefined because the radicand cannot be negative.

$$\left(\left(-2\right)^{2}-6\left(-2\right)-16\right)\sqrt{-2-3}=0$$

Substitute 3 for x in the equation \left(x^{2}-6x-16\right)\sqrt{x-3}=0.

$$\left(3^{2}-6\times 3-16\right)\sqrt{3-3}=0$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

Substitute 8 for x in the equation \left(x^{2}-6x-16\right)\sqrt{x-3}=0.

$$\left(8^{2}-6\times 8-16\right)\sqrt{8-3}=0$$

Simplify. The value x=8 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{x-3}x^{2}=6\sqrt{x-3}x+16\sqrt{x-3}.

$$x=3 x=8$$

Solve for x (complex solution)

$$x=-2$$

$$x=8$$

$$x=3$$