Solve sqrt{x+5}+1=6/sqrt{x+5}

$$\sqrt { x + 5 } + 1 = \frac { 6 } { \sqrt { x + 5 } }$$

Solve for x

$$x=-1$$

Solution Steps

Subtract 1 from both sides of the equation.

$$\sqrt{x+5}=\frac{6}{\sqrt{x+5}}-1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\sqrt{x+5}}{\sqrt{x+5}}.

$$\sqrt{x+5}=\frac{6}{\sqrt{x+5}}-\frac{\sqrt{x+5}}{\sqrt{x+5}}$$

Since \frac{6}{\sqrt{x+5}} and \frac{\sqrt{x+5}}{\sqrt{x+5}} have the same denominator, subtract them by subtracting their numerators.

$$\sqrt{x+5}=\frac{6-\sqrt{x+5}}{\sqrt{x+5}}$$

Square both sides of the equation.

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

Calculate \sqrt{x+5} to the power of 2 and get x+5.

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

To raise \frac{6-\sqrt{x+5}}{\sqrt{x+5}} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-\sqrt{x+5}\right)^{2}.

$$x+5=\frac{36-12\sqrt{x+5}+\left(\sqrt{x+5}\right)^{2}}{\left(\sqrt{x+5}\right)^{2}}$$

Calculate \sqrt{x+5} to the power of 2 and get x+5.

$$x+5=\frac{36-12\sqrt{x+5}+x+5}{\left(\sqrt{x+5}\right)^{2}}$$

Add 36 and 5 to get 41.

$$x+5=\frac{41-12\sqrt{x+5}+x}{\left(\sqrt{x+5}\right)^{2}}$$

Calculate \sqrt{x+5} to the power of 2 and get x+5.

$$x+5=\frac{41-12\sqrt{x+5}+x}{x+5}$$

Multiply both sides of the equation by x+5.

$$\left(x+5\right)x+\left(x+5\right)\times 5=41-12\sqrt{x+5}+x$$

Subtract 41+x from both sides of the equation.

$$\left(x+5\right)x+\left(x+5\right)\times 5-\left(41+x\right)=-12\sqrt{x+5}$$

Use the distributive property to multiply x+5 by x.

$$x^{2}+5x+\left(x+5\right)\times 5-\left(41+x\right)=-12\sqrt{x+5}$$

Use the distributive property to multiply x+5 by 5.

$$x^{2}+5x+5x+25-\left(41+x\right)=-12\sqrt{x+5}$$

Combine 5x and 5x to get 10x.

$$x^{2}+10x+25-\left(41+x\right)=-12\sqrt{x+5}$$

To find the opposite of 41+x, find the opposite of each term.

$$x^{2}+10x+25-41-x=-12\sqrt{x+5}$$

Subtract 41 from 25 to get -16.

$$x^{2}+10x-16-x=-12\sqrt{x+5}$$

Combine 10x and -x to get 9x.

$$x^{2}+9x-16=-12\sqrt{x+5}$$

Square both sides of the equation.

$$\left(x^{2}+9x-16\right)^{2}=\left(-12\sqrt{x+5}\right)^{2}$$

Square x^{2}+9x-16.

$$x^{4}+18x^{3}+49x^{2}-288x+256=\left(-12\sqrt{x+5}\right)^{2}$$

Expand \left(-12\sqrt{x+5}\right)^{2}.

$$x^{4}+18x^{3}+49x^{2}-288x+256=\left(-12\right)^{2}\left(\sqrt{x+5}\right)^{2}$$

Calculate -12 to the power of 2 and get 144.

$$x^{4}+18x^{3}+49x^{2}-288x+256=144\left(\sqrt{x+5}\right)^{2}$$

Calculate \sqrt{x+5} to the power of 2 and get x+5.

$$x^{4}+18x^{3}+49x^{2}-288x+256=144\left(x+5\right)$$

Use the distributive property to multiply 144 by x+5.

$$x^{4}+18x^{3}+49x^{2}-288x+256=144x+720$$

Subtract 144x from both sides.

$$x^{4}+18x^{3}+49x^{2}-288x+256-144x=720$$

Combine -288x and -144x to get -432x.

$$x^{4}+18x^{3}+49x^{2}-432x+256=720$$

Subtract 720 from both sides.

$$x^{4}+18x^{3}+49x^{2}-432x+256-720=0$$

Subtract 720 from 256 to get -464.

$$x^{4}+18x^{3}+49x^{2}-432x-464=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -464 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±464,±232,±116,±58,±29,±16,±8,±4,±2,±1$$

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^{4}+18x^{3}+49x^{2}-432x-464 by x+1 to get x^{3}+17x^{2}+32x-464. Solve the equation where the result equals to 0.

$$x^{3}+17x^{2}+32x-464=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -464 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±464,±232,±116,±58,±29,±16,±8,±4,±2,±1$$

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}+17x^{2}+32x-464 by x-4 to get x^{2}+21x+116. Solve the equation where the result equals to 0.

$$x^{2}+21x+116=0$$

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, 21 for b, and 116 for c in the quadratic formula.

$$x=\frac{-21±\sqrt{21^{2}-4\times 1\times 116}}{2}$$

Do the calculations.

$$x=\frac{-21±\sqrt{-23}}{2}$$

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.

$$x=-1 x=4$$

Substitute -1 for x in the equation \sqrt{x+5}+1=\frac{6}{\sqrt{x+5}}.

$$\sqrt{-1+5}+1=\frac{6}{\sqrt{-1+5}}$$

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

$$3=3$$

Substitute 4 for x in the equation \sqrt{x+5}+1=\frac{6}{\sqrt{x+5}}.

$$\sqrt{4+5}+1=\frac{6}{\sqrt{4+5}}$$

Simplify. The value x=4 does not satisfy the equation.

$$4=2$$

Substitute -1 for x in the equation \sqrt{x+5}+1=\frac{6}{\sqrt{x+5}}.

$$\sqrt{-1+5}+1=\frac{6}{\sqrt{-1+5}}$$

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

$$3=3$$

Equation \sqrt{x+5}=\frac{-\sqrt{x+5}+6}{\sqrt{x+5}} has a unique solution.

$$x=-1$$