Solve 2x^2+1/2x-x^2/x+1=3x^2-1/x(x+1)

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

Solve for x

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

$$x=1$$

Steps Using Factoring By Grouping

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

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

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

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

Use the distributive property to multiply 2 by 3x^{2}-1.

$$2x^{3}+x+2x^{2}+1-2x^{3}=6x^{2}-2$$

Subtract 6x^{2} from both sides.

$$2x^{3}+x+2x^{2}+1-2x^{3}-6x^{2}=-2$$

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

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

Add 2 to both sides.

$$2x^{3}+x-4x^{2}+1-2x^{3}+2=0$$

Add 1 and 2 to get 3.

$$2x^{3}+x-4x^{2}+3-2x^{3}=0$$

Combine 2x^{3} and -2x^{3} to get 0.

$$x-4x^{2}+3=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-4x^{2}+x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=1 ab=-4\times 3=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=4 b=-3$$

Rewrite -4x^{2}+x+3 as \left(-4x^{2}+4x\right)+\left(-3x+3\right).

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

Factor out 4x in the first and 3 in the second group.

$$4x\left(-x+1\right)+3\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(4x+3\right)$$

To find equation solutions, solve -x+1=0 and 4x+3=0.

$$x=1 x=-\frac{3}{4}$$

Steps Using the Quadratic Formula

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

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

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

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

Use the distributive property to multiply 2 by 3x^{2}-1.

$$2x^{3}+x+2x^{2}+1-2x^{3}=6x^{2}-2$$

Subtract 6x^{2} from both sides.

$$2x^{3}+x+2x^{2}+1-2x^{3}-6x^{2}=-2$$

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

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

Add 2 to both sides.

$$2x^{3}+x-4x^{2}+1-2x^{3}+2=0$$

Add 1 and 2 to get 3.

$$2x^{3}+x-4x^{2}+3-2x^{3}=0$$

Combine 2x^{3} and -2x^{3} to get 0.

$$x-4x^{2}+3=0$$

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.

$$-4x^{2}+x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 1 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-4\right)\times 3}}{2\left(-4\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-4\right)\times 3}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-1±\sqrt{1+16\times 3}}{2\left(-4\right)}$$

Multiply 16 times 3.

$$x=\frac{-1±\sqrt{1+48}}{2\left(-4\right)}$$

Add 1 to 48.

$$x=\frac{-1±\sqrt{49}}{2\left(-4\right)}$$

Take the square root of 49.

$$x=\frac{-1±7}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-1±7}{-8}$$

Now solve the equation x=\frac{-1±7}{-8} when ± is plus. Add -1 to 7.

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

Reduce the fraction \frac{6}{-8} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-1±7}{-8} when ± is minus. Subtract 7 from -1.

$$x=-\frac{8}{-8}$$

Divide -8 by -8.

$$x=1$$

The equation is now solved.

$$x=-\frac{3}{4} x=1$$

Steps for Completing the Square

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

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

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

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

Use the distributive property to multiply 2 by 3x^{2}-1.

$$2x^{3}+x+2x^{2}+1-2x^{3}=6x^{2}-2$$

Subtract 6x^{2} from both sides.

$$2x^{3}+x+2x^{2}+1-2x^{3}-6x^{2}=-2$$

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

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

Subtract 1 from both sides.

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

Subtract 1 from -2 to get -3.

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

Combine 2x^{3} and -2x^{3} to get 0.

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

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.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Divide 1 by -4.

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

Divide -3 by -4.

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

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\frac{3}{4}+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{3}{4}+\frac{1}{64}$$

Add \frac{3}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{49}{64}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{49}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{7}{8} x-\frac{1}{8}=-\frac{7}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$x=1 x=-\frac{3}{4}$$