Solve sqrt{x+3}+3/sqrt{x+3}=4

$$\sqrt { x + 3 } + \frac { 3 } { \sqrt { x + 3 } } = 4$$

Solve for x

$$x=6$$

$$x=-2$$

Solution Steps

Subtract \frac{3}{\sqrt{x+3}} from both sides of the equation.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{\sqrt{x+3}}{\sqrt{x+3}}.

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

Since \frac{4\sqrt{x+3}}{\sqrt{x+3}} and \frac{3}{\sqrt{x+3}} have the same denominator, subtract them by subtracting their numerators.

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

Square both sides of the equation.

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

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

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

To raise \frac{4\sqrt{x+3}-3}{\sqrt{x+3}} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4\sqrt{x+3}-3\right)^{2}.

$$x+3=\frac{16\left(\sqrt{x+3}\right)^{2}-24\sqrt{x+3}+9}{\left(\sqrt{x+3}\right)^{2}}$$

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

$$x+3=\frac{16\left(x+3\right)-24\sqrt{x+3}+9}{\left(\sqrt{x+3}\right)^{2}}$$

Use the distributive property to multiply 16 by x+3.

$$x+3=\frac{16x+48-24\sqrt{x+3}+9}{\left(\sqrt{x+3}\right)^{2}}$$

Add 48 and 9 to get 57.

$$x+3=\frac{16x+57-24\sqrt{x+3}}{\left(\sqrt{x+3}\right)^{2}}$$

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

$$x+3=\frac{16x+57-24\sqrt{x+3}}{x+3}$$

Multiply both sides of the equation by x+3.

$$\left(x+3\right)x+\left(x+3\right)\times 3=16x+57-24\sqrt{x+3}$$

Subtract 16x+57 from both sides of the equation.

$$\left(x+3\right)x+\left(x+3\right)\times 3-\left(16x+57\right)=-24\sqrt{x+3}$$

Use the distributive property to multiply x+3 by x.

$$x^{2}+3x+\left(x+3\right)\times 3-\left(16x+57\right)=-24\sqrt{x+3}$$

Use the distributive property to multiply x+3 by 3.

$$x^{2}+3x+3x+9-\left(16x+57\right)=-24\sqrt{x+3}$$

Combine 3x and 3x to get 6x.

$$x^{2}+6x+9-\left(16x+57\right)=-24\sqrt{x+3}$$

To find the opposite of 16x+57, find the opposite of each term.

$$x^{2}+6x+9-16x-57=-24\sqrt{x+3}$$

Combine 6x and -16x to get -10x.

$$x^{2}-10x+9-57=-24\sqrt{x+3}$$

Subtract 57 from 9 to get -48.

$$x^{2}-10x-48=-24\sqrt{x+3}$$

Square both sides of the equation.

$$\left(x^{2}-10x-48\right)^{2}=\left(-24\sqrt{x+3}\right)^{2}$$

Square x^{2}-10x-48.

$$x^{4}-20x^{3}+4x^{2}+960x+2304=\left(-24\sqrt{x+3}\right)^{2}$$

Expand \left(-24\sqrt{x+3}\right)^{2}.

$$x^{4}-20x^{3}+4x^{2}+960x+2304=\left(-24\right)^{2}\left(\sqrt{x+3}\right)^{2}$$

Calculate -24 to the power of 2 and get 576.

$$x^{4}-20x^{3}+4x^{2}+960x+2304=576\left(\sqrt{x+3}\right)^{2}$$

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

$$x^{4}-20x^{3}+4x^{2}+960x+2304=576\left(x+3\right)$$

Use the distributive property to multiply 576 by x+3.

$$x^{4}-20x^{3}+4x^{2}+960x+2304=576x+1728$$

Subtract 576x from both sides.

$$x^{4}-20x^{3}+4x^{2}+960x+2304-576x=1728$$

Combine 960x and -576x to get 384x.

$$x^{4}-20x^{3}+4x^{2}+384x+2304=1728$$

Subtract 1728 from both sides.

$$x^{4}-20x^{3}+4x^{2}+384x+2304-1728=0$$

Subtract 1728 from 2304 to get 576.

$$x^{4}-20x^{3}+4x^{2}+384x+576=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 576 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±576,±288,±192,±144,±96,±72,±64,±48,±36,±32,±24,±18,±16,±12,±9,±8,±6,±4,±3,±2,±1$$

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^{4}-20x^{3}+4x^{2}+384x+576 by x+2 to get x^{3}-22x^{2}+48x+288. Solve the equation where the result equals to 0.

$$x^{3}-22x^{2}+48x+288=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 288 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±288,±144,±96,±72,±48,±36,±32,±24,±18,±16,±12,±9,±8,±6,±4,±3,±2,±1$$

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.

$$x=6$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-22x^{2}+48x+288 by x-6 to get x^{2}-16x-48. Solve the equation where the result equals to 0.

$$x^{2}-16x-48=0$$

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, -16 for b, and -48 for c in the quadratic formula.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 1\left(-48\right)}}{2}$$

Do the calculations.

$$x=\frac{16±8\sqrt{7}}{2}$$

Solve the equation x^{2}-16x-48=0 when ± is plus and when ± is minus.

$$x=8-4\sqrt{7} x=4\sqrt{7}+8$$

List all found solutions.

$$x=-2 x=6 x=8-4\sqrt{7} x=4\sqrt{7}+8$$

Substitute -2 for x in the equation \sqrt{x+3}+\frac{3}{\sqrt{x+3}}=4.

$$\sqrt{-2+3}+\frac{3}{\sqrt{-2+3}}=4$$

Simplify. The value x=-2 satisfies the equation.

$$4=4$$

Substitute 6 for x in the equation \sqrt{x+3}+\frac{3}{\sqrt{x+3}}=4.

$$\sqrt{6+3}+\frac{3}{\sqrt{6+3}}=4$$

Simplify. The value x=6 satisfies the equation.

$$4=4$$

Substitute 8-4\sqrt{7} for x in the equation \sqrt{x+3}+\frac{3}{\sqrt{x+3}}=4.

$$\sqrt{8-4\sqrt{7}+3}+\frac{3}{\sqrt{8-4\sqrt{7}+3}}=4$$

Simplify. The value x=8-4\sqrt{7} does not satisfy the equation.

$$2\times 7^{\frac{1}{2}}=4$$

Substitute 4\sqrt{7}+8 for x in the equation \sqrt{x+3}+\frac{3}{\sqrt{x+3}}=4.

$$\sqrt{4\sqrt{7}+8+3}+\frac{3}{\sqrt{4\sqrt{7}+8+3}}=4$$

Simplify. The value x=4\sqrt{7}+8 does not satisfy the equation.

$$2\times 7^{\frac{1}{2}}=4$$

Substitute -2 for x in the equation \sqrt{x+3}+\frac{3}{\sqrt{x+3}}=4.

$$\sqrt{-2+3}+\frac{3}{\sqrt{-2+3}}=4$$

Simplify. The value x=-2 satisfies the equation.

$$4=4$$

Substitute 6 for x in the equation \sqrt{x+3}+\frac{3}{\sqrt{x+3}}=4.

$$\sqrt{6+3}+\frac{3}{\sqrt{6+3}}=4$$

Simplify. The value x=6 satisfies the equation.

$$4=4$$

List all solutions of \sqrt{x+3}=\frac{4\sqrt{x+3}-3}{\sqrt{x+3}}.

$$x=-2 x=6$$