Solve frac{sqrt{x+1}(x^2-16)}{-x^2-9}=0

$$\frac { \sqrt { x + 1 } ( x ^ { 2 } - 16 ) } { - x ^ { 2 } - 9 } = 0$$

Solve for x

$$x=4$$

$$x=-1$$

Solution Steps

Use the distributive property to multiply \sqrt{x+1} by x^{2}-16.

$$\frac{\sqrt{x+1}x^{2}-16\sqrt{x+1}}{-x^{2}-9}=0$$

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

$$\sqrt{x+1}x^{2}-16\sqrt{x+1}=0$$

Subtract -16\sqrt{x+1} from both sides of the equation.

$$\sqrt{x+1}x^{2}=16\sqrt{x+1}$$

Square both sides of the equation.

$$\left(\sqrt{x+1}x^{2}\right)^{2}=\left(16\sqrt{x+1}\right)^{2}$$

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

$$\left(\sqrt{x+1}\right)^{2}\left(x^{2}\right)^{2}=\left(16\sqrt{x+1}\right)^{2}$$

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

$$\left(\sqrt{x+1}\right)^{2}x^{4}=\left(16\sqrt{x+1}\right)^{2}$$

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

$$\left(x+1\right)x^{4}=\left(16\sqrt{x+1}\right)^{2}$$

Use the distributive property to multiply x+1 by x^{4}.

$$x^{5}+x^{4}=\left(16\sqrt{x+1}\right)^{2}$$

Expand \left(16\sqrt{x+1}\right)^{2}.

$$x^{5}+x^{4}=16^{2}\left(\sqrt{x+1}\right)^{2}$$

Calculate 16 to the power of 2 and get 256.

$$x^{5}+x^{4}=256\left(\sqrt{x+1}\right)^{2}$$

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

$$x^{5}+x^{4}=256\left(x+1\right)$$

Use the distributive property to multiply 256 by x+1.

$$x^{5}+x^{4}=256x+256$$

Subtract 256x from both sides.

$$x^{5}+x^{4}-256x=256$$

Subtract 256 from both sides.

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

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{5}+x^{4}-256x-256 by x+1 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.

$$x=-1 x=4 x=-4$$

Substitute -1 for x in the equation \frac{\sqrt{x+1}\left(x^{2}-16\right)}{-x^{2}-9}=0.

$$\frac{\sqrt{-1+1}\left(\left(-1\right)^{2}-16\right)}{-\left(-1\right)^{2}-9}=0$$

Simplify. The value x=-1 satisfies the equation.

$$0=0$$

Substitute 4 for x in the equation \frac{\sqrt{x+1}\left(x^{2}-16\right)}{-x^{2}-9}=0.

$$\frac{\sqrt{4+1}\left(4^{2}-16\right)}{-4^{2}-9}=0$$

Simplify. The value x=4 satisfies the equation.

$$0=0$$

Substitute -4 for x in the equation \frac{\sqrt{x+1}\left(x^{2}-16\right)}{-x^{2}-9}=0. The expression \sqrt{-4+1} is undefined because the radicand cannot be negative.

$$\frac{\sqrt{-4+1}\left(\left(-4\right)^{2}-16\right)}{-\left(-4\right)^{2}-9}=0$$

List all solutions of \sqrt{x+1}x^{2}=16\sqrt{x+1}.

$$x=-1 x=4$$

Solve for x (complex solution)

$$x=-4$$

$$x=4$$

$$x=-1$$