Solve x^2+(x+2)^2/x(x+2)=34/15

$$\frac { x ^ { 2 } + ( x + 2 ) ^ { 2 } } { x ( x + 2 ) } = \frac { 34 } { 15 }$$

Solve for x

$$x=-5$$

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by 15x\left(x+2\right), the least common multiple of x\left(x+2\right),15.

$$15\left(x^{2}+\left(x+2\right)^{2}\right)=34x\left(x+2\right)$$

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

$$15\left(x^{2}+x^{2}+4x+4\right)=34x\left(x+2\right)$$

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

$$15\left(2x^{2}+4x+4\right)=34x\left(x+2\right)$$

Use the distributive property to multiply 15 by 2x^{2}+4x+4.

$$30x^{2}+60x+60=34x\left(x+2\right)$$

Use the distributive property to multiply 34x by x+2.

$$30x^{2}+60x+60=34x^{2}+68x$$

Subtract 34x^{2} from both sides.

$$30x^{2}+60x+60-34x^{2}=68x$$

Combine 30x^{2} and -34x^{2} to get -4x^{2}.

$$-4x^{2}+60x+60=68x$$

Subtract 68x from both sides.

$$-4x^{2}+60x+60-68x=0$$

Combine 60x and -68x to get -8x.

$$-4x^{2}-8x+60=0$$

Divide both sides by 4.

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

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

$$a+b=-2 ab=-15=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=3 b=-5$$

Rewrite -x^{2}-2x+15 as \left(-x^{2}+3x\right)+\left(-5x+15\right).

$$\left(-x^{2}+3x\right)+\left(-5x+15\right)$$

Factor out x in the first and 5 in the second group.

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

Factor out common term -x+3 by using distributive property.

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

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

$$x=3 x=-5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by 15x\left(x+2\right), the least common multiple of x\left(x+2\right),15.

$$15\left(x^{2}+\left(x+2\right)^{2}\right)=34x\left(x+2\right)$$

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

$$15\left(x^{2}+x^{2}+4x+4\right)=34x\left(x+2\right)$$

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

$$15\left(2x^{2}+4x+4\right)=34x\left(x+2\right)$$

Use the distributive property to multiply 15 by 2x^{2}+4x+4.

$$30x^{2}+60x+60=34x\left(x+2\right)$$

Use the distributive property to multiply 34x by x+2.

$$30x^{2}+60x+60=34x^{2}+68x$$

Subtract 34x^{2} from both sides.

$$30x^{2}+60x+60-34x^{2}=68x$$

Combine 30x^{2} and -34x^{2} to get -4x^{2}.

$$-4x^{2}+60x+60=68x$$

Subtract 68x from both sides.

$$-4x^{2}+60x+60-68x=0$$

Combine 60x and -68x to get -8x.

$$-4x^{2}-8x+60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -8 for b, and 60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-4\right)\times 60}}{2\left(-4\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-4\right)\times 60}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-8\right)±\sqrt{64+16\times 60}}{2\left(-4\right)}$$

Multiply 16 times 60.

$$x=\frac{-\left(-8\right)±\sqrt{64+960}}{2\left(-4\right)}$$

Add 64 to 960.

$$x=\frac{-\left(-8\right)±\sqrt{1024}}{2\left(-4\right)}$$

Take the square root of 1024.

$$x=\frac{-\left(-8\right)±32}{2\left(-4\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±32}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{8±32}{-8}$$

Now solve the equation x=\frac{8±32}{-8} when ± is plus. Add 8 to 32.

$$x=\frac{40}{-8}$$

Divide 40 by -8.

$$x=-5$$

Now solve the equation x=\frac{8±32}{-8} when ± is minus. Subtract 32 from 8.

$$x=-\frac{24}{-8}$$

Divide -24 by -8.

$$x=3$$

The equation is now solved.

$$x=-5 x=3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by 15x\left(x+2\right), the least common multiple of x\left(x+2\right),15.

$$15\left(x^{2}+\left(x+2\right)^{2}\right)=34x\left(x+2\right)$$

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

$$15\left(x^{2}+x^{2}+4x+4\right)=34x\left(x+2\right)$$

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

$$15\left(2x^{2}+4x+4\right)=34x\left(x+2\right)$$

Use the distributive property to multiply 15 by 2x^{2}+4x+4.

$$30x^{2}+60x+60=34x\left(x+2\right)$$

Use the distributive property to multiply 34x by x+2.

$$30x^{2}+60x+60=34x^{2}+68x$$

Subtract 34x^{2} from both sides.

$$30x^{2}+60x+60-34x^{2}=68x$$

Combine 30x^{2} and -34x^{2} to get -4x^{2}.

$$-4x^{2}+60x+60=68x$$

Subtract 68x from both sides.

$$-4x^{2}+60x+60-68x=0$$

Combine 60x and -68x to get -8x.

$$-4x^{2}-8x+60=0$$

Subtract 60 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}-8x=-60$$

Divide both sides by -4.

$$\frac{-4x^{2}-8x}{-4}=-\frac{60}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{8}{-4}\right)x=-\frac{60}{-4}$$

Divide -8 by -4.

$$x^{2}+2x=-\frac{60}{-4}$$

Divide -60 by -4.

$$x^{2}+2x=15$$

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.

$$x^{2}+2x+1^{2}=15+1^{2}$$

Square 1.

$$x^{2}+2x+1=15+1$$

Add 15 to 1.

$$x^{2}+2x+1=16$$

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}.

$$\left(x+1\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{16}$$

Simplify.

$$x+1=4 x+1=-4$$

Subtract 1 from both sides of the equation.

$$x=3 x=-5$$