Solve 1/2+-2c/c-2=9/4c

$$\frac { 1 } { 2 } + \frac { - 2 c } { c - 2 } = \frac { 9 } { 4 } c$$

Solve for c

$$c=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

Variable c cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 4\left(c-2\right), the least common multiple of 2,c-2,4.

$$4\left(c-2\right)\times \frac{1}{2}+4\left(-2\right)c=\frac{9}{4}c\times 4\left(c-2\right)$$

Use the distributive property to multiply 4 by c-2.

$$\left(4c-8\right)\times \frac{1}{2}+4\left(-2\right)c=\frac{9}{4}c\times 4\left(c-2\right)$$

Use the distributive property to multiply 4c-8 by \frac{1}{2}.

$$2c-4+4\left(-2\right)c=\frac{9}{4}c\times 4\left(c-2\right)$$

Multiply 4 and -2 to get -8.

$$2c-4-8c=\frac{9}{4}c\times 4\left(c-2\right)$$

Combine 2c and -8c to get -6c.

$$-6c-4=\frac{9}{4}c\times 4\left(c-2\right)$$

Multiply \frac{9}{4} and 4 to get 9.

$$-6c-4=9c\left(c-2\right)$$

Use the distributive property to multiply 9c by c-2.

$$-6c-4=9c^{2}-18c$$

Subtract 9c^{2} from both sides.

$$-6c-4-9c^{2}=-18c$$

Add 18c to both sides.

$$-6c-4-9c^{2}+18c=0$$

Combine -6c and 18c to get 12c.

$$12c-4-9c^{2}=0$$

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

$$-9c^{2}+12c-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9c^{2}+ac+bc-4. To find a and b, set up a system to be solved.

$$a+b=12 ab=-9\left(-4\right)=36$$

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 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 12.

$$a=6 b=6$$

Rewrite -9c^{2}+12c-4 as \left(-9c^{2}+6c\right)+\left(6c-4\right).

$$\left(-9c^{2}+6c\right)+\left(6c-4\right)$$

Factor out -3c in the first and 2 in the second group.

$$-3c\left(3c-2\right)+2\left(3c-2\right)$$

Factor out common term 3c-2 by using distributive property.

$$\left(3c-2\right)\left(-3c+2\right)$$

To find equation solutions, solve 3c-2=0 and -3c+2=0.

$$c=\frac{2}{3} c=\frac{2}{3}$$

Steps Using the Quadratic Formula

Variable c cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 4\left(c-2\right), the least common multiple of 2,c-2,4.

$$4\left(c-2\right)\times \frac{1}{2}+4\left(-2\right)c=\frac{9}{4}c\times 4\left(c-2\right)$$

Use the distributive property to multiply 4 by c-2.

$$\left(4c-8\right)\times \frac{1}{2}+4\left(-2\right)c=\frac{9}{4}c\times 4\left(c-2\right)$$

Use the distributive property to multiply 4c-8 by \frac{1}{2}.

$$2c-4+4\left(-2\right)c=\frac{9}{4}c\times 4\left(c-2\right)$$

Multiply 4 and -2 to get -8.

$$2c-4-8c=\frac{9}{4}c\times 4\left(c-2\right)$$

Combine 2c and -8c to get -6c.

$$-6c-4=\frac{9}{4}c\times 4\left(c-2\right)$$

Multiply \frac{9}{4} and 4 to get 9.

$$-6c-4=9c\left(c-2\right)$$

Use the distributive property to multiply 9c by c-2.

$$-6c-4=9c^{2}-18c$$

Subtract 9c^{2} from both sides.

$$-6c-4-9c^{2}=-18c$$

Add 18c to both sides.

$$-6c-4-9c^{2}+18c=0$$

Combine -6c and 18c to get 12c.

$$12c-4-9c^{2}=0$$

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.

$$-9c^{2}+12c-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 12 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-12±\sqrt{12^{2}-4\left(-9\right)\left(-4\right)}}{2\left(-9\right)}$$

Square 12.

$$c=\frac{-12±\sqrt{144-4\left(-9\right)\left(-4\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$c=\frac{-12±\sqrt{144+36\left(-4\right)}}{2\left(-9\right)}$$

Multiply 36 times -4.

$$c=\frac{-12±\sqrt{144-144}}{2\left(-9\right)}$$

Add 144 to -144.

$$c=\frac{-12±\sqrt{0}}{2\left(-9\right)}$$

Take the square root of 0.

$$c=-\frac{12}{2\left(-9\right)}$$

Multiply 2 times -9.

$$c=-\frac{12}{-18}$$

Reduce the fraction \frac{-12}{-18} to lowest terms by extracting and canceling out 6.

$$c=\frac{2}{3}$$

Steps for Completing the Square

Variable c cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 4\left(c-2\right), the least common multiple of 2,c-2,4.

$$4\left(c-2\right)\times \frac{1}{2}+4\left(-2\right)c=\frac{9}{4}c\times 4\left(c-2\right)$$

Use the distributive property to multiply 4 by c-2.

$$\left(4c-8\right)\times \frac{1}{2}+4\left(-2\right)c=\frac{9}{4}c\times 4\left(c-2\right)$$

Use the distributive property to multiply 4c-8 by \frac{1}{2}.

$$2c-4+4\left(-2\right)c=\frac{9}{4}c\times 4\left(c-2\right)$$

Multiply 4 and -2 to get -8.

$$2c-4-8c=\frac{9}{4}c\times 4\left(c-2\right)$$

Combine 2c and -8c to get -6c.

$$-6c-4=\frac{9}{4}c\times 4\left(c-2\right)$$

Multiply \frac{9}{4} and 4 to get 9.

$$-6c-4=9c\left(c-2\right)$$

Use the distributive property to multiply 9c by c-2.

$$-6c-4=9c^{2}-18c$$

Subtract 9c^{2} from both sides.

$$-6c-4-9c^{2}=-18c$$

Add 18c to both sides.

$$-6c-4-9c^{2}+18c=0$$

Combine -6c and 18c to get 12c.

$$12c-4-9c^{2}=0$$

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

$$12c-9c^{2}=4$$

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.

$$-9c^{2}+12c=4$$

Divide both sides by -9.

$$\frac{-9c^{2}+12c}{-9}=\frac{4}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$c^{2}+\frac{12}{-9}c=\frac{4}{-9}$$

Reduce the fraction \frac{12}{-9} to lowest terms by extracting and canceling out 3.

$$c^{2}-\frac{4}{3}c=\frac{4}{-9}$$

Divide 4 by -9.

$$c^{2}-\frac{4}{3}c=-\frac{4}{9}$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}-\frac{4}{3}c+\left(-\frac{2}{3}\right)^{2}=-\frac{4}{9}+\left(-\frac{2}{3}\right)^{2}$$

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$c^{2}-\frac{4}{3}c+\frac{4}{9}=\frac{-4+4}{9}$$

Add -\frac{4}{9} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$c^{2}-\frac{4}{3}c+\frac{4}{9}=0$$

Factor c^{2}-\frac{4}{3}c+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(c-\frac{2}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c-\frac{2}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$c-\frac{2}{3}=0 c-\frac{2}{3}=0$$

Add \frac{2}{3} to both sides of the equation.

$$c=\frac{2}{3} c=\frac{2}{3}$$

The equation is now solved. Solutions are the same.

$$c=\frac{2}{3}$$