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

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

Solve for x

$$x = \frac{\sqrt{7} + 1}{2} \approx 1.822875656$$

$$x=\frac{1-\sqrt{7}}{2}\approx -0.822875656$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,-\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(x+2\right)\left(2x+1\right), the least common multiple of 2x+1,x+2.

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

Use the distributive property to multiply 2x+1 by 3.

$$x+2+6x+3=\left(x+2\right)\left(2x+1\right)$$

Combine x and 6x to get 7x.

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

Add 2 and 3 to get 5.

$$7x+5=\left(x+2\right)\left(2x+1\right)$$

Use the distributive property to multiply x+2 by 2x+1 and combine like terms.

$$7x+5=2x^{2}+5x+2$$

Subtract 2x^{2} from both sides.

$$7x+5-2x^{2}=5x+2$$

Subtract 5x from both sides.

$$7x+5-2x^{2}-5x=2$$

Combine 7x and -5x to get 2x.

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

Subtract 2 from both sides.

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

Subtract 2 from 5 to get 3.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -2.

$$x=\frac{-2±\sqrt{4+8\times 3}}{2\left(-2\right)}$$

Multiply 8 times 3.

$$x=\frac{-2±\sqrt{4+24}}{2\left(-2\right)}$$

Add 4 to 24.

$$x=\frac{-2±\sqrt{28}}{2\left(-2\right)}$$

Take the square root of 28.

$$x=\frac{-2±2\sqrt{7}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-2±2\sqrt{7}}{-4}$$

Now solve the equation x=\frac{-2±2\sqrt{7}}{-4} when ± is plus. Add -2 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-2}{-4}$$

Divide -2+2\sqrt{7} by -4.

$$x=\frac{1-\sqrt{7}}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{7}}{-4} when ± is minus. Subtract 2\sqrt{7} from -2.

$$x=\frac{-2\sqrt{7}-2}{-4}$$

Divide -2-2\sqrt{7} by -4.

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

The equation is now solved.

$$x=\frac{1-\sqrt{7}}{2} x=\frac{\sqrt{7}+1}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,-\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(x+2\right)\left(2x+1\right), the least common multiple of 2x+1,x+2.

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

Use the distributive property to multiply 2x+1 by 3.

$$x+2+6x+3=\left(x+2\right)\left(2x+1\right)$$

Combine x and 6x to get 7x.

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

Add 2 and 3 to get 5.

$$7x+5=\left(x+2\right)\left(2x+1\right)$$

Use the distributive property to multiply x+2 by 2x+1 and combine like terms.

$$7x+5=2x^{2}+5x+2$$

Subtract 2x^{2} from both sides.

$$7x+5-2x^{2}=5x+2$$

Subtract 5x from both sides.

$$7x+5-2x^{2}-5x=2$$

Combine 7x and -5x to get 2x.

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

Subtract 5 from both sides.

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

Subtract 5 from 2 to get -3.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -2.

$$\frac{-2x^{2}+2x}{-2}=-\frac{3}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{2}{-2}x=-\frac{3}{-2}$$

Divide 2 by -2.

$$x^{2}-x=-\frac{3}{-2}$$

Divide -3 by -2.

$$x^{2}-x=\frac{3}{2}$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{3}{2}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=\frac{3}{2}+\frac{1}{4}$$

Add \frac{3}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{7}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{7}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{7}}{2} x-\frac{1}{2}=-\frac{\sqrt{7}}{2}$$

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

$$x=\frac{\sqrt{7}+1}{2} x=\frac{1-\sqrt{7}}{2}$$