Solve 2x+3/x-1-5/x=6

$$\frac { 2 x + 3 } { x - 1 } - \frac { 5 } { x } = 6$$

Solve for x

$$x = \frac{\sqrt{6} + 1}{2} \approx 1.724744871$$

$$x=\frac{1-\sqrt{6}}{2}\approx -0.724744871$$

Steps Using the Quadratic Formula

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

$$x\left(2x+3\right)-\left(x-1\right)\times 5=6x\left(x-1\right)$$

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

$$2x^{2}+3x-\left(x-1\right)\times 5=6x\left(x-1\right)$$

Use the distributive property to multiply x-1 by 5.

$$2x^{2}+3x-\left(5x-5\right)=6x\left(x-1\right)$$

To find the opposite of 5x-5, find the opposite of each term.

$$2x^{2}+3x-5x+5=6x\left(x-1\right)$$

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

$$2x^{2}-2x+5=6x\left(x-1\right)$$

Use the distributive property to multiply 6x by x-1.

$$2x^{2}-2x+5=6x^{2}-6x$$

Subtract 6x^{2} from both sides.

$$2x^{2}-2x+5-6x^{2}=-6x$$

Combine 2x^{2} and -6x^{2} to get -4x^{2}.

$$-4x^{2}-2x+5=-6x$$

Add 6x to both sides.

$$-4x^{2}-2x+5+6x=0$$

Combine -2x and 6x to get 4x.

$$-4x^{2}+4x+5=0$$

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-4\right)\times 5}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-4±\sqrt{16+16\times 5}}{2\left(-4\right)}$$

Multiply 16 times 5.

$$x=\frac{-4±\sqrt{16+80}}{2\left(-4\right)}$$

Add 16 to 80.

$$x=\frac{-4±\sqrt{96}}{2\left(-4\right)}$$

Take the square root of 96.

$$x=\frac{-4±4\sqrt{6}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-4±4\sqrt{6}}{-8}$$

Now solve the equation x=\frac{-4±4\sqrt{6}}{-8} when ± is plus. Add -4 to 4\sqrt{6}.

$$x=\frac{4\sqrt{6}-4}{-8}$$

Divide -4+4\sqrt{6} by -8.

$$x=\frac{1-\sqrt{6}}{2}$$

Now solve the equation x=\frac{-4±4\sqrt{6}}{-8} when ± is minus. Subtract 4\sqrt{6} from -4.

$$x=\frac{-4\sqrt{6}-4}{-8}$$

Divide -4-4\sqrt{6} by -8.

$$x=\frac{\sqrt{6}+1}{2}$$

The equation is now solved.

$$x=\frac{1-\sqrt{6}}{2} x=\frac{\sqrt{6}+1}{2}$$

Steps for Completing the Square

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

$$x\left(2x+3\right)-\left(x-1\right)\times 5=6x\left(x-1\right)$$

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

$$2x^{2}+3x-\left(x-1\right)\times 5=6x\left(x-1\right)$$

Use the distributive property to multiply x-1 by 5.

$$2x^{2}+3x-\left(5x-5\right)=6x\left(x-1\right)$$

To find the opposite of 5x-5, find the opposite of each term.

$$2x^{2}+3x-5x+5=6x\left(x-1\right)$$

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

$$2x^{2}-2x+5=6x\left(x-1\right)$$

Use the distributive property to multiply 6x by x-1.

$$2x^{2}-2x+5=6x^{2}-6x$$

Subtract 6x^{2} from both sides.

$$2x^{2}-2x+5-6x^{2}=-6x$$

Combine 2x^{2} and -6x^{2} to get -4x^{2}.

$$-4x^{2}-2x+5=-6x$$

Add 6x to both sides.

$$-4x^{2}-2x+5+6x=0$$

Combine -2x and 6x to get 4x.

$$-4x^{2}+4x+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}+4x=-5$$

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Divide 4 by -4.

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

Divide -5 by -4.

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

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.

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

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

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

Add \frac{5}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{6}}{2} x-\frac{1}{2}=-\frac{\sqrt{6}}{2}$$

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

$$x=\frac{\sqrt{6}+1}{2} x=\frac{1-\sqrt{6}}{2}$$