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

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

Solve for x

$$x=\frac{\sqrt{21}-5}{2}\approx -0.208712153$$

$$x=\frac{-\sqrt{21}-5}{2}\approx -4.791287847$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{1}{2},0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x+1\right), the least common multiple of x,2x+1.

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

Consider \left(2x+1\right)\left(2x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

To find the opposite of x^{2}+3x, find the opposite of each term.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

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

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

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

Subtract 4x^{2} from both sides.

$$3x^{2}-1-3x-4x^{2}=2x$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}-1-3x=2x$$

Subtract 2x from both sides.

$$-x^{2}-1-3x-2x=0$$

Combine -3x and -2x to get -5x.

$$-x^{2}-1-5x=0$$

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.

$$-x^{2}-5x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\left(-1\right)}}{2\left(-1\right)}$$

Multiply 4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25-4}}{2\left(-1\right)}$$

Add 25 to -4.

$$x=\frac{-\left(-5\right)±\sqrt{21}}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{21}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±\sqrt{21}}{-2}$$

Now solve the equation x=\frac{5±\sqrt{21}}{-2} when ± is plus. Add 5 to \sqrt{21}.

$$x=\frac{\sqrt{21}+5}{-2}$$

Divide 5+\sqrt{21} by -2.

$$x=\frac{-\sqrt{21}-5}{2}$$

Now solve the equation x=\frac{5±\sqrt{21}}{-2} when ± is minus. Subtract \sqrt{21} from 5.

$$x=\frac{5-\sqrt{21}}{-2}$$

Divide 5-\sqrt{21} by -2.

$$x=\frac{\sqrt{21}-5}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{21}-5}{2} x=\frac{\sqrt{21}-5}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{1}{2},0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x+1\right), the least common multiple of x,2x+1.

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

Consider \left(2x+1\right)\left(2x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

To find the opposite of x^{2}+3x, find the opposite of each term.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

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

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

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

Subtract 4x^{2} from both sides.

$$3x^{2}-1-3x-4x^{2}=2x$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}-1-3x=2x$$

Subtract 2x from both sides.

$$-x^{2}-1-3x-2x=0$$

Combine -3x and -2x to get -5x.

$$-x^{2}-1-5x=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$-x^{2}-5x=1$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -5 by -1.

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

Divide 1 by -1.

$$x^{2}+5x=-1$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=-1+\frac{25}{4}$$

Add -1 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{21}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{21}}{2} x+\frac{5}{2}=-\frac{\sqrt{21}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{21}-5}{2} x=\frac{-\sqrt{21}-5}{2}$$