Solve sqrt{12x}=4x+6/3

$$\sqrt { 12 x } = \frac { 4 x + 6 } { 3 }$$

Solve for x

$$x=3$$

$$x=\frac{3}{4}=0.75$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{12x}\right)^{2}=\left(\frac{4x+6}{3}\right)^{2}$$

Calculate \sqrt{12x} to the power of 2 and get 12x.

$$12x=\left(\frac{4x+6}{3}\right)^{2}$$

To raise \frac{4x+6}{3} to a power, raise both numerator and denominator to the power and then divide.

$$12x=\frac{\left(4x+6\right)^{2}}{3^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+6\right)^{2}.

$$12x=\frac{16x^{2}+48x+36}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$12x=\frac{16x^{2}+48x+36}{9}$$

Divide each term of 16x^{2}+48x+36 by 9 to get \frac{16}{9}x^{2}+\frac{16}{3}x+4.

$$12x=\frac{16}{9}x^{2}+\frac{16}{3}x+4$$

Subtract \frac{16}{9}x^{2} from both sides.

$$12x-\frac{16}{9}x^{2}=\frac{16}{3}x+4$$

Subtract \frac{16}{3}x from both sides.

$$12x-\frac{16}{9}x^{2}-\frac{16}{3}x=4$$

Combine 12x and -\frac{16}{3}x to get \frac{20}{3}x.

$$\frac{20}{3}x-\frac{16}{9}x^{2}=4$$

Subtract 4 from both sides.

$$\frac{20}{3}x-\frac{16}{9}x^{2}-4=0$$

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.

$$-\frac{16}{9}x^{2}+\frac{20}{3}x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{16}{9} for a, \frac{20}{3} for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{20}{3}±\sqrt{\left(\frac{20}{3}\right)^{2}-4\left(-\frac{16}{9}\right)\left(-4\right)}}{2\left(-\frac{16}{9}\right)}$$

Square \frac{20}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{20}{3}±\sqrt{\frac{400}{9}-4\left(-\frac{16}{9}\right)\left(-4\right)}}{2\left(-\frac{16}{9}\right)}$$

Multiply -4 times -\frac{16}{9}.

$$x=\frac{-\frac{20}{3}±\sqrt{\frac{400}{9}+\frac{64}{9}\left(-4\right)}}{2\left(-\frac{16}{9}\right)}$$

Multiply \frac{64}{9} times -4.

$$x=\frac{-\frac{20}{3}±\sqrt{\frac{400-256}{9}}}{2\left(-\frac{16}{9}\right)}$$

Add \frac{400}{9} to -\frac{256}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{20}{3}±\sqrt{16}}{2\left(-\frac{16}{9}\right)}$$

Take the square root of 16.

$$x=\frac{-\frac{20}{3}±4}{2\left(-\frac{16}{9}\right)}$$

Multiply 2 times -\frac{16}{9}.

$$x=\frac{-\frac{20}{3}±4}{-\frac{32}{9}}$$

Now solve the equation x=\frac{-\frac{20}{3}±4}{-\frac{32}{9}} when ± is plus. Add -\frac{20}{3} to 4.

$$x=-\frac{\frac{8}{3}}{-\frac{32}{9}}$$

Divide -\frac{8}{3} by -\frac{32}{9} by multiplying -\frac{8}{3} by the reciprocal of -\frac{32}{9}.

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

Now solve the equation x=\frac{-\frac{20}{3}±4}{-\frac{32}{9}} when ± is minus. Subtract 4 from -\frac{20}{3}.

$$x=-\frac{\frac{32}{3}}{-\frac{32}{9}}$$

Divide -\frac{32}{3} by -\frac{32}{9} by multiplying -\frac{32}{3} by the reciprocal of -\frac{32}{9}.

$$x=3$$

The equation is now solved.

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

Substitute \frac{3}{4} for x in the equation \sqrt{12x}=\frac{4x+6}{3}.

$$\sqrt{12\times \frac{3}{4}}=\frac{4\times \frac{3}{4}+6}{3}$$

Simplify. The value x=\frac{3}{4} satisfies the equation.

$$3=3$$

Substitute 3 for x in the equation \sqrt{12x}=\frac{4x+6}{3}.

$$\sqrt{12\times 3}=\frac{4\times 3+6}{3}$$

Simplify. The value x=3 satisfies the equation.

$$6=6$$

List all solutions of \sqrt{12x}=\frac{4x+6}{3}.

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