Solve 3x+2+1/3x+2=7

$$3 x + 2 + \frac { 1 } { 3 x + 2 } = 7$$

Solve for x

$$x=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

$$x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+2.

$$3x\left(3x+2\right)+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

Use the distributive property to multiply 3x by 3x+2.

$$9x^{2}+6x+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

Use the distributive property to multiply 3x+2 by 2.

$$9x^{2}+6x+6x+4+1=7\left(3x+2\right)$$

Combine 6x and 6x to get 12x.

$$9x^{2}+12x+4+1=7\left(3x+2\right)$$

Add 4 and 1 to get 5.

$$9x^{2}+12x+5=7\left(3x+2\right)$$

Use the distributive property to multiply 7 by 3x+2.

$$9x^{2}+12x+5=21x+14$$

Subtract 21x from both sides.

$$9x^{2}+12x+5-21x=14$$

Combine 12x and -21x to get -9x.

$$9x^{2}-9x+5=14$$

Subtract 14 from both sides.

$$9x^{2}-9x+5-14=0$$

Subtract 14 from 5 to get -9.

$$9x^{2}-9x-9=0$$

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

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

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 9\left(-9\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-9\right)±\sqrt{81-36\left(-9\right)}}{2\times 9}$$

Multiply -36 times -9.

$$x=\frac{-\left(-9\right)±\sqrt{81+324}}{2\times 9}$$

Add 81 to 324.

$$x=\frac{-\left(-9\right)±\sqrt{405}}{2\times 9}$$

Take the square root of 405.

$$x=\frac{-\left(-9\right)±9\sqrt{5}}{2\times 9}$$

The opposite of -9 is 9.

$$x=\frac{9±9\sqrt{5}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{9±9\sqrt{5}}{18}$$

Now solve the equation x=\frac{9±9\sqrt{5}}{18} when ± is plus. Add 9 to 9\sqrt{5}.

$$x=\frac{9\sqrt{5}+9}{18}$$

Divide 9+9\sqrt{5} by 18.

$$x=\frac{\sqrt{5}+1}{2}$$

Now solve the equation x=\frac{9±9\sqrt{5}}{18} when ± is minus. Subtract 9\sqrt{5} from 9.

$$x=\frac{9-9\sqrt{5}}{18}$$

Divide 9-9\sqrt{5} by 18.

$$x=\frac{1-\sqrt{5}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$

Steps for Completing the Square

Variable x cannot be equal to -\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+2.

$$3x\left(3x+2\right)+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

Use the distributive property to multiply 3x by 3x+2.

$$9x^{2}+6x+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

Use the distributive property to multiply 3x+2 by 2.

$$9x^{2}+6x+6x+4+1=7\left(3x+2\right)$$

Combine 6x and 6x to get 12x.

$$9x^{2}+12x+4+1=7\left(3x+2\right)$$

Add 4 and 1 to get 5.

$$9x^{2}+12x+5=7\left(3x+2\right)$$

Use the distributive property to multiply 7 by 3x+2.

$$9x^{2}+12x+5=21x+14$$

Subtract 21x from both sides.

$$9x^{2}+12x+5-21x=14$$

Combine 12x and -21x to get -9x.

$$9x^{2}-9x+5=14$$

Subtract 5 from both sides.

$$9x^{2}-9x=14-5$$

Subtract 5 from 14 to get 9.

$$9x^{2}-9x=9$$

Divide both sides by 9.

$$\frac{9x^{2}-9x}{9}=\frac{9}{9}$$

Dividing by 9 undoes the multiplication by 9.

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

Divide -9 by 9.

$$x^{2}-x=\frac{9}{9}$$

Divide 9 by 9.

$$x^{2}-x=1$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 1 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{5}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{5}}{2} x-\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$