Solve 2x+9/2x-6=(x+3/x-1)^2

$$\frac { 2 x + 9 } { 2 x - 6 } = ( \frac { x + 3 } { x - 1 } ) ^ { 2 }$$

Solve for x

$$x=-7$$

$$x=9$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right).

$$2x+9=2\left(x-3\right)\times \left(\frac{x+3}{x-1}\right)^{2}$$

To raise \frac{x+3}{x-1} to a power, raise both numerator and denominator to the power and then divide.

$$2x+9=2\left(x-3\right)\times \frac{\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Express 2\times \frac{\left(x+3\right)^{2}}{\left(x-1\right)^{2}} as a single fraction.

$$2x+9=\frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}\left(x-3\right)$$

Use the distributive property to multiply \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}} by x-3.

$$2x+9=\frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}x-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)}{\left(x-1\right)^{2}}x-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}x-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Express \frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}x as a single fraction.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}-3\times \frac{2\left(x^{2}+6x+9\right)}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}-3\times \frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}$$

Express -3\times \frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1} as a single fraction.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}+\frac{-3\times 2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}$$

Since \frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1} and \frac{-3\times 2\left(x^{2}+6x+9\right)}{x^{2}-2x+1} have the same denominator, add them by adding their numerators.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x-3\times 2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}$$

Do the multiplications in 2\left(x^{2}+6x+9\right)x-3\times 2\left(x^{2}+6x+9\right).

$$2x+9=\frac{2x^{3}+12x^{2}+18x-6x^{2}-36x-54}{x^{2}-2x+1}$$

Combine like terms in 2x^{3}+12x^{2}+18x-6x^{2}-36x-54.

$$2x+9=\frac{2x^{3}+6x^{2}-18x-54}{x^{2}-2x+1}$$

Subtract \frac{2x^{3}+6x^{2}-18x-54}{x^{2}-2x+1} from both sides.

$$2x+9-\frac{2x^{3}+6x^{2}-18x-54}{x^{2}-2x+1}=0$$

Factor x^{2}-2x+1.

$$2x+9-\frac{2x^{3}+6x^{2}-18x-54}{\left(x-1\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x+9 times \frac{\left(x-1\right)^{2}}{\left(x-1\right)^{2}}.

$$\frac{\left(2x+9\right)\left(x-1\right)^{2}}{\left(x-1\right)^{2}}-\frac{2x^{3}+6x^{2}-18x-54}{\left(x-1\right)^{2}}=0$$

Since \frac{\left(2x+9\right)\left(x-1\right)^{2}}{\left(x-1\right)^{2}} and \frac{2x^{3}+6x^{2}-18x-54}{\left(x-1\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(2x+9\right)\left(x-1\right)^{2}-\left(2x^{3}+6x^{2}-18x-54\right)}{\left(x-1\right)^{2}}=0$$

Do the multiplications in \left(2x+9\right)\left(x-1\right)^{2}-\left(2x^{3}+6x^{2}-18x-54\right).

$$\frac{2x^{3}-4x^{2}+2x+9x^{2}-18x+9-2x^{3}-6x^{2}+18x+54}{\left(x-1\right)^{2}}=0$$

Combine like terms in 2x^{3}-4x^{2}+2x+9x^{2}-18x+9-2x^{3}-6x^{2}+18x+54.

$$\frac{-x^{2}+2x+63}{\left(x-1\right)^{2}}=0$$

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)^{2}.

$$-x^{2}+2x+63=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+63. To find a and b, set up a system to be solved.

$$a+b=2 ab=-63=-63$$

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 -63.

$$-1,63 -3,21 -7,9$$

Calculate the sum for each pair.

$$-1+63=62 -3+21=18 -7+9=2$$

The solution is the pair that gives sum 2.

$$a=9 b=-7$$

Rewrite -x^{2}+2x+63 as \left(-x^{2}+9x\right)+\left(-7x+63\right).

$$\left(-x^{2}+9x\right)+\left(-7x+63\right)$$

Factor out -x in the first and -7 in the second group.

$$-x\left(x-9\right)-7\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x-7\right)$$

To find equation solutions, solve x-9=0 and -x-7=0.

$$x=9 x=-7$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right).

$$2x+9=2\left(x-3\right)\times \left(\frac{x+3}{x-1}\right)^{2}$$

To raise \frac{x+3}{x-1} to a power, raise both numerator and denominator to the power and then divide.

$$2x+9=2\left(x-3\right)\times \frac{\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Express 2\times \frac{\left(x+3\right)^{2}}{\left(x-1\right)^{2}} as a single fraction.

$$2x+9=\frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}\left(x-3\right)$$

Use the distributive property to multiply \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}} by x-3.

$$2x+9=\frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}x-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)}{\left(x-1\right)^{2}}x-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}x-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Express \frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}x as a single fraction.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}-3\times \frac{2\left(x^{2}+6x+9\right)}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}-3\times \frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}$$

Express -3\times \frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1} as a single fraction.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}+\frac{-3\times 2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}$$

Since \frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1} and \frac{-3\times 2\left(x^{2}+6x+9\right)}{x^{2}-2x+1} have the same denominator, add them by adding their numerators.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x-3\times 2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}$$

Do the multiplications in 2\left(x^{2}+6x+9\right)x-3\times 2\left(x^{2}+6x+9\right).

$$2x+9=\frac{2x^{3}+12x^{2}+18x-6x^{2}-36x-54}{x^{2}-2x+1}$$

Combine like terms in 2x^{3}+12x^{2}+18x-6x^{2}-36x-54.

$$2x+9=\frac{2x^{3}+6x^{2}-18x-54}{x^{2}-2x+1}$$

Subtract \frac{2x^{3}+6x^{2}-18x-54}{x^{2}-2x+1} from both sides.

$$2x+9-\frac{2x^{3}+6x^{2}-18x-54}{x^{2}-2x+1}=0$$

Factor x^{2}-2x+1.

$$2x+9-\frac{2x^{3}+6x^{2}-18x-54}{\left(x-1\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x+9 times \frac{\left(x-1\right)^{2}}{\left(x-1\right)^{2}}.

$$\frac{\left(2x+9\right)\left(x-1\right)^{2}}{\left(x-1\right)^{2}}-\frac{2x^{3}+6x^{2}-18x-54}{\left(x-1\right)^{2}}=0$$

Since \frac{\left(2x+9\right)\left(x-1\right)^{2}}{\left(x-1\right)^{2}} and \frac{2x^{3}+6x^{2}-18x-54}{\left(x-1\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(2x+9\right)\left(x-1\right)^{2}-\left(2x^{3}+6x^{2}-18x-54\right)}{\left(x-1\right)^{2}}=0$$

Do the multiplications in \left(2x+9\right)\left(x-1\right)^{2}-\left(2x^{3}+6x^{2}-18x-54\right).

$$\frac{2x^{3}-4x^{2}+2x+9x^{2}-18x+9-2x^{3}-6x^{2}+18x+54}{\left(x-1\right)^{2}}=0$$

Combine like terms in 2x^{3}-4x^{2}+2x+9x^{2}-18x+9-2x^{3}-6x^{2}+18x+54.

$$\frac{-x^{2}+2x+63}{\left(x-1\right)^{2}}=0$$

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)^{2}.

$$-x^{2}+2x+63=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 2 for b, and 63 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-1\right)\times 63}}{2\left(-1\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-1\right)\times 63}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-2±\sqrt{4+4\times 63}}{2\left(-1\right)}$$

Multiply 4 times 63.

$$x=\frac{-2±\sqrt{4+252}}{2\left(-1\right)}$$

Add 4 to 252.

$$x=\frac{-2±\sqrt{256}}{2\left(-1\right)}$$

Take the square root of 256.

$$x=\frac{-2±16}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-2±16}{-2}$$

Now solve the equation x=\frac{-2±16}{-2} when ± is plus. Add -2 to 16.

$$x=\frac{14}{-2}$$

Divide 14 by -2.

$$x=-7$$

Now solve the equation x=\frac{-2±16}{-2} when ± is minus. Subtract 16 from -2.

$$x=-\frac{18}{-2}$$

Divide -18 by -2.

$$x=9$$

The equation is now solved.

$$x=-7 x=9$$

Steps for Completing the Square

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right).

$$2x+9=2\left(x-3\right)\times \left(\frac{x+3}{x-1}\right)^{2}$$

To raise \frac{x+3}{x-1} to a power, raise both numerator and denominator to the power and then divide.

$$2x+9=2\left(x-3\right)\times \frac{\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Express 2\times \frac{\left(x+3\right)^{2}}{\left(x-1\right)^{2}} as a single fraction.

$$2x+9=\frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}\left(x-3\right)$$

Use the distributive property to multiply \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}} by x-3.

$$2x+9=\frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}x-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)}{\left(x-1\right)^{2}}x-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}x-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Express \frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}x as a single fraction.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}-3\times \frac{2\left(x+3\right)^{2}}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}-3\times \frac{2\left(x^{2}+6x+9\right)}{\left(x-1\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}-3\times \frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}$$

Express -3\times \frac{2\left(x^{2}+6x+9\right)}{x^{2}-2x+1} as a single fraction.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1}+\frac{-3\times 2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}$$

Since \frac{2\left(x^{2}+6x+9\right)x}{x^{2}-2x+1} and \frac{-3\times 2\left(x^{2}+6x+9\right)}{x^{2}-2x+1} have the same denominator, add them by adding their numerators.

$$2x+9=\frac{2\left(x^{2}+6x+9\right)x-3\times 2\left(x^{2}+6x+9\right)}{x^{2}-2x+1}$$

Do the multiplications in 2\left(x^{2}+6x+9\right)x-3\times 2\left(x^{2}+6x+9\right).

$$2x+9=\frac{2x^{3}+12x^{2}+18x-6x^{2}-36x-54}{x^{2}-2x+1}$$

Combine like terms in 2x^{3}+12x^{2}+18x-6x^{2}-36x-54.

$$2x+9=\frac{2x^{3}+6x^{2}-18x-54}{x^{2}-2x+1}$$

Subtract \frac{2x^{3}+6x^{2}-18x-54}{x^{2}-2x+1} from both sides.

$$2x+9-\frac{2x^{3}+6x^{2}-18x-54}{x^{2}-2x+1}=0$$

Factor x^{2}-2x+1.

$$2x+9-\frac{2x^{3}+6x^{2}-18x-54}{\left(x-1\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x+9 times \frac{\left(x-1\right)^{2}}{\left(x-1\right)^{2}}.

$$\frac{\left(2x+9\right)\left(x-1\right)^{2}}{\left(x-1\right)^{2}}-\frac{2x^{3}+6x^{2}-18x-54}{\left(x-1\right)^{2}}=0$$

Since \frac{\left(2x+9\right)\left(x-1\right)^{2}}{\left(x-1\right)^{2}} and \frac{2x^{3}+6x^{2}-18x-54}{\left(x-1\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(2x+9\right)\left(x-1\right)^{2}-\left(2x^{3}+6x^{2}-18x-54\right)}{\left(x-1\right)^{2}}=0$$

Do the multiplications in \left(2x+9\right)\left(x-1\right)^{2}-\left(2x^{3}+6x^{2}-18x-54\right).

$$\frac{2x^{3}-4x^{2}+2x+9x^{2}-18x+9-2x^{3}-6x^{2}+18x+54}{\left(x-1\right)^{2}}=0$$

Combine like terms in 2x^{3}-4x^{2}+2x+9x^{2}-18x+9-2x^{3}-6x^{2}+18x+54.

$$\frac{-x^{2}+2x+63}{\left(x-1\right)^{2}}=0$$

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)^{2}.

$$-x^{2}+2x+63=0$$

Subtract 63 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}+2x=-63$$

Divide both sides by -1.

$$\frac{-x^{2}+2x}{-1}=-\frac{63}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{2}{-1}x=-\frac{63}{-1}$$

Divide 2 by -1.

$$x^{2}-2x=-\frac{63}{-1}$$

Divide -63 by -1.

$$x^{2}-2x=63$$

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

$$x^{2}-2x+1=63+1$$

Add 63 to 1.

$$x^{2}-2x+1=64$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=8 x-1=-8$$

Add 1 to both sides of the equation.

$$x=9 x=-7$$