Solve quad2/x=2-3/x-2

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

Solve for x

$$x=4$$

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

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

$$\left(x-2\right)\times 2=x\left(x-2\right)\times 2-x\times 3$$

Use the distributive property to multiply x-2 by 2.

$$2x-4=x\left(x-2\right)\times 2-x\times 3$$

Use the distributive property to multiply x by x-2.

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

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

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

Subtract 2x^{2} from both sides.

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

Add 4x to both sides.

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

Combine 2x and 4x to get 6x.

$$6x-4-2x^{2}=-x\times 3$$

Add x\times 3 to both sides.

$$6x-4-2x^{2}+x\times 3=0$$

Combine 6x and x\times 3 to get 9x.

$$9x-4-2x^{2}=0$$

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

$$-2x^{2}+9x-4=0$$

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

$$a+b=9 ab=-2\left(-4\right)=8$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 9.

$$a=8 b=1$$

Rewrite -2x^{2}+9x-4 as \left(-2x^{2}+8x\right)+\left(x-4\right).

$$\left(-2x^{2}+8x\right)+\left(x-4\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(-x+4\right)-\left(-x+4\right)$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(2x-1\right)$$

To find equation solutions, solve -x+4=0 and 2x-1=0.

$$x=4 x=\frac{1}{2}$$

Steps Using the Quadratic Formula

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

$$\left(x-2\right)\times 2=x\left(x-2\right)\times 2-x\times 3$$

Use the distributive property to multiply x-2 by 2.

$$2x-4=x\left(x-2\right)\times 2-x\times 3$$

Use the distributive property to multiply x by x-2.

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

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

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

Subtract 2x^{2} from both sides.

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

Add 4x to both sides.

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

Combine 2x and 4x to get 6x.

$$6x-4-2x^{2}=-x\times 3$$

Add x\times 3 to both sides.

$$6x-4-2x^{2}+x\times 3=0$$

Combine 6x and x\times 3 to get 9x.

$$9x-4-2x^{2}=0$$

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.

$$-2x^{2}+9x-4=0$$

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

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

Square 9.

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

Multiply -4 times -2.

$$x=\frac{-9±\sqrt{81+8\left(-4\right)}}{2\left(-2\right)}$$

Multiply 8 times -4.

$$x=\frac{-9±\sqrt{81-32}}{2\left(-2\right)}$$

Add 81 to -32.

$$x=\frac{-9±\sqrt{49}}{2\left(-2\right)}$$

Take the square root of 49.

$$x=\frac{-9±7}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-9±7}{-4}$$

Now solve the equation x=\frac{-9±7}{-4} when ± is plus. Add -9 to 7.

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

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-9±7}{-4} when ± is minus. Subtract 7 from -9.

$$x=-\frac{16}{-4}$$

Divide -16 by -4.

$$x=4$$

The equation is now solved.

$$x=\frac{1}{2} x=4$$

Steps for Completing the Square

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

$$\left(x-2\right)\times 2=x\left(x-2\right)\times 2-x\times 3$$

Use the distributive property to multiply x-2 by 2.

$$2x-4=x\left(x-2\right)\times 2-x\times 3$$

Use the distributive property to multiply x by x-2.

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

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

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

Subtract 2x^{2} from both sides.

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

Add 4x to both sides.

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

Combine 2x and 4x to get 6x.

$$6x-4-2x^{2}=-x\times 3$$

Add x\times 3 to both sides.

$$6x-4-2x^{2}+x\times 3=0$$

Combine 6x and x\times 3 to get 9x.

$$9x-4-2x^{2}=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$9x-2x^{2}=4$$

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.

$$-2x^{2}+9x=4$$

Divide both sides by -2.

$$\frac{-2x^{2}+9x}{-2}=\frac{4}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{9}{-2}x=\frac{4}{-2}$$

Divide 9 by -2.

$$x^{2}-\frac{9}{2}x=\frac{4}{-2}$$

Divide 4 by -2.

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

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=-2+\frac{81}{16}$$

Add -2 to \frac{81}{16}.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{49}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{7}{4} x-\frac{9}{4}=-\frac{7}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=4 x=\frac{1}{2}$$