Solve sqrt{3x}+sqrt{12}=x+5/sqrt{3}

$$\sqrt { 3 x } + \sqrt { 12 } = \frac { x + 5 } { \sqrt { 3 } }$$

Solve for x

$$x = \frac{3 \sqrt{13} + 11}{2} \approx 10.908326913$$

Solution Steps

Subtract \sqrt{12} from both sides of the equation.

$$\sqrt{3x}=\frac{x+5}{\sqrt{3}}-\sqrt{12}$$

Rationalize the denominator of \frac{x+5}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\sqrt{3x}=\frac{\left(x+5\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\sqrt{12}$$

The square of \sqrt{3} is 3.

$$\sqrt{3x}=\frac{\left(x+5\right)\sqrt{3}}{3}-\sqrt{12}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\sqrt{3x}=\frac{\left(x+5\right)\sqrt{3}}{3}-2\sqrt{3}$$

Use the distributive property to multiply x+5 by \sqrt{3}.

$$\sqrt{3x}=\frac{x\sqrt{3}+5\sqrt{3}}{3}-2\sqrt{3}$$

Multiply both sides of the equation by 3.

$$3\sqrt{3x}=x\sqrt{3}+5\sqrt{3}-6\sqrt{3}$$

Reorder the terms.

$$3\sqrt{3x}=\sqrt{3}x-6\sqrt{3}+5\sqrt{3}$$

Square both sides of the equation.

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

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

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

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{3x}\right)^{2}=\left(\sqrt{3}x-6\sqrt{3}+5\sqrt{3}\right)^{2}$$

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

$$9\times 3x=\left(\sqrt{3}x-6\sqrt{3}+5\sqrt{3}\right)^{2}$$

Multiply 9 and 3 to get 27.

$$27x=\left(\sqrt{3}x-6\sqrt{3}+5\sqrt{3}\right)^{2}$$

Combine -6\sqrt{3} and 5\sqrt{3} to get -\sqrt{3}.

$$27x=\left(\sqrt{3}x-\sqrt{3}\right)^{2}$$

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

$$27x=\left(\sqrt{3}\right)^{2}x^{2}-2\sqrt{3}x\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$27x=\left(\sqrt{3}\right)^{2}x^{2}-2\times 3x+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$27x=3x^{2}-2\times 3x+\left(\sqrt{3}\right)^{2}$$

Multiply -2 and 3 to get -6.

$$27x=3x^{2}-6x+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$27x=3x^{2}-6x+3$$

Subtract 3x^{2} from both sides.

$$27x-3x^{2}=-6x+3$$

Add 6x to both sides.

$$27x-3x^{2}+6x=3$$

Combine 27x and 6x to get 33x.

$$33x-3x^{2}=3$$

Subtract 3 from both sides.

$$33x-3x^{2}-3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3x^{2}+33x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 33 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-33±\sqrt{33^{2}-4\left(-3\right)\left(-3\right)}}{2\left(-3\right)}$$

Square 33.

$$x=\frac{-33±\sqrt{1089-4\left(-3\right)\left(-3\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-33±\sqrt{1089+12\left(-3\right)}}{2\left(-3\right)}$$

Multiply 12 times -3.

$$x=\frac{-33±\sqrt{1089-36}}{2\left(-3\right)}$$

Add 1089 to -36.

$$x=\frac{-33±\sqrt{1053}}{2\left(-3\right)}$$

Take the square root of 1053.

$$x=\frac{-33±9\sqrt{13}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-33±9\sqrt{13}}{-6}$$

Now solve the equation x=\frac{-33±9\sqrt{13}}{-6} when ± is plus. Add -33 to 9\sqrt{13}.

$$x=\frac{9\sqrt{13}-33}{-6}$$

Divide -33+9\sqrt{13} by -6.

$$x=\frac{11-3\sqrt{13}}{2}$$

Now solve the equation x=\frac{-33±9\sqrt{13}}{-6} when ± is minus. Subtract 9\sqrt{13} from -33.

$$x=\frac{-9\sqrt{13}-33}{-6}$$

Divide -33-9\sqrt{13} by -6.

$$x=\frac{3\sqrt{13}+11}{2}$$

The equation is now solved.

$$x=\frac{11-3\sqrt{13}}{2} x=\frac{3\sqrt{13}+11}{2}$$

Substitute \frac{11-3\sqrt{13}}{2} for x in the equation \sqrt{3x}+\sqrt{12}=\frac{x+5}{\sqrt{3}}.

$$\sqrt{3\times \frac{11-3\sqrt{13}}{2}}+\sqrt{12}=\frac{\frac{11-3\sqrt{13}}{2}+5}{\sqrt{3}}$$

Simplify. The value x=\frac{11-3\sqrt{13}}{2} does not satisfy the equation.

$$\frac{1}{2}\times 39^{\frac{1}{2}}+\frac{1}{2}\times 3^{\frac{1}{2}}=\frac{1}{3}\left(\frac{21}{2}-\frac{3}{2}\times 13^{\frac{1}{2}}\right)\times 3^{\frac{1}{2}}$$

Substitute \frac{3\sqrt{13}+11}{2} for x in the equation \sqrt{3x}+\sqrt{12}=\frac{x+5}{\sqrt{3}}.

$$\sqrt{3\times \frac{3\sqrt{13}+11}{2}}+\sqrt{12}=\frac{\frac{3\sqrt{13}+11}{2}+5}{\sqrt{3}}$$

Simplify. The value x=\frac{3\sqrt{13}+11}{2} satisfies the equation.

$$\frac{1}{2}\times 39^{\frac{1}{2}}+\frac{7}{2}\times 3^{\frac{1}{2}}=\frac{1}{3}\left(\frac{3}{2}\times 13^{\frac{1}{2}}+\frac{21}{2}\right)\times 3^{\frac{1}{2}}$$

Equation 3\sqrt{3x}=\sqrt{3}x-6\sqrt{3}+5\sqrt{3} has a unique solution.

$$x=\frac{3\sqrt{13}+11}{2}$$