Solve x^2-6x/x^2+2x+x^2+3x/x^2+2x=sqrt{2x-3/x+2}

$$\frac { x ^ { 2 } - 6 x } { x ^ { 2 } + 2 x } + \frac { x ^ { 2 } + 3 x } { x ^ { 2 } + 2 x } = \sqrt { \frac { 2 x - 3 } { x + 2 } }$$

Solve for x

$$x=5$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

Square both sides of the equation.

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

Factor the expressions that are not already factored in \frac{x^{2}-6x}{x^{2}+2x}.

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

Cancel out x in both numerator and denominator.

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

Factor the expressions that are not already factored in \frac{x^{2}+3x}{x^{2}+2x}.

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

Cancel out x in both numerator and denominator.

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

Since \frac{x-6}{x+2} and \frac{x+3}{x+2} have the same denominator, add them by adding their numerators.

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

Combine like terms in x-6+x+3.

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

To raise \frac{2x-3}{x+2} to a power, raise both numerator and denominator to the power and then divide.

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

Calculate \sqrt{\frac{2x-3}{x+2}} to the power of 2 and get \frac{2x-3}{x+2}.

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

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

$$\frac{4x^{2}-12x+9}{\left(x+2\right)^{2}}=\frac{2x-3}{x+2}$$

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

$$\frac{4x^{2}-12x+9}{x^{2}+4x+4}=\frac{2x-3}{x+2}$$

Multiply both sides of the equation by \left(x+2\right)^{2}, the least common multiple of x^{2}+4x+4,x+2.

$$4x^{2}-12x+9=\left(x+2\right)\left(2x-3\right)$$

Use the distributive property to multiply x+2 by 2x-3 and combine like terms.

$$4x^{2}-12x+9=2x^{2}+x-6$$

Subtract 2x^{2} from both sides.

$$4x^{2}-12x+9-2x^{2}=x-6$$

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

$$2x^{2}-12x+9=x-6$$

Subtract x from both sides.

$$2x^{2}-12x+9-x=-6$$

Combine -12x and -x to get -13x.

$$2x^{2}-13x+9=-6$$

Add 6 to both sides.

$$2x^{2}-13x+9+6=0$$

Add 9 and 6 to get 15.

$$2x^{2}-13x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-13 ab=2\times 15=30$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -13.

$$a=-10 b=-3$$

Rewrite 2x^{2}-13x+15 as \left(2x^{2}-10x\right)+\left(-3x+15\right).

$$\left(2x^{2}-10x\right)+\left(-3x+15\right)$$

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

$$2x\left(x-5\right)-3\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(2x-3\right)$$

To find equation solutions, solve x-5=0 and 2x-3=0.

$$x=5 x=\frac{3}{2}$$

Substitute 5 for x in the equation \frac{x^{2}-6x}{x^{2}+2x}+\frac{x^{2}+3x}{x^{2}+2x}=\sqrt{\frac{2x-3}{x+2}}.

$$\frac{5^{2}-6\times 5}{5^{2}+2\times 5}+\frac{5^{2}+3\times 5}{5^{2}+2\times 5}=\sqrt{\frac{2\times 5-3}{5+2}}$$

Simplify. The value x=5 satisfies the equation.

$$1=1$$

Substitute \frac{3}{2} for x in the equation \frac{x^{2}-6x}{x^{2}+2x}+\frac{x^{2}+3x}{x^{2}+2x}=\sqrt{\frac{2x-3}{x+2}}.

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

Simplify. The value x=\frac{3}{2} satisfies the equation.

$$0=0$$

List all solutions of \frac{x^{2}+x^{2}+3x-6x}{x^{2}+2x}=\sqrt{\frac{2x-3}{x+2}}.

$$x=5 x=\frac{3}{2}$$