Solve sqrt{x+4}=1/x+4

$$\sqrt{ x+4 } = \frac{ 1 }{ x+4 }$$

Solve for x

$$x=-3$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x+4} to the power of 2 and get x+4.

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

To raise \frac{1}{x+4} to a power, raise both numerator and denominator to the power and then divide.

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

Calculate 1 to the power of 2 and get 1.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x+4=\frac{1}{x^{2}+8x+16}$$

Multiply both sides of the equation by \left(x+4\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$\left(x^{2}+8x+16\right)x+\left(x+4\right)^{2}\times 4=1$$

Use the distributive property to multiply x^{2}+8x+16 by x.

$$x^{3}+8x^{2}+16x+\left(x+4\right)^{2}\times 4=1$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x^{3}+8x^{2}+16x+\left(x^{2}+8x+16\right)\times 4=1$$

Use the distributive property to multiply x^{2}+8x+16 by 4.

$$x^{3}+8x^{2}+16x+4x^{2}+32x+64=1$$

Combine 8x^{2} and 4x^{2} to get 12x^{2}.

$$x^{3}+12x^{2}+16x+32x+64=1$$

Combine 16x and 32x to get 48x.

$$x^{3}+12x^{2}+48x+64=1$$

Subtract 1 from both sides.

$$x^{3}+12x^{2}+48x+64-1=0$$

Subtract 1 from 64 to get 63.

$$x^{3}+12x^{2}+48x+63=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 63 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±63,±21,±9,±7,±3,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+12x^{2}+48x+63 by x+3 to get x^{2}+9x+21. Solve the equation where the result equals to 0.

$$x^{2}+9x+21=0$$

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, 9 for b, and 21 for c in the quadratic formula.

$$x=\frac{-9±\sqrt{9^{2}-4\times 1\times 21}}{2}$$

Do the calculations.

$$x=\frac{-9±\sqrt{-3}}{2}$$

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.

$$x=-3$$

Substitute -3 for x in the equation \sqrt{x+4}=\frac{1}{x+4}.

$$\sqrt{-3+4}=\frac{1}{-3+4}$$

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

$$1=1$$

Equation \sqrt{x+4}=\frac{1}{x+4} has a unique solution.

$$x=-3$$