Solve x^2+5x/x+1+x+1/1=10

$$\frac { x ^ { 2 } + 5 x } { x + 1 } + \frac { x + 1 } { 1 } = 10$$

Solve for x

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

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

$$x^{2}+5x+\left(x+1\right)\left(x+1\right)=10\left(x+1\right)$$

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

$$x^{2}+5x+\left(x+1\right)^{2}=10\left(x+1\right)$$

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

$$x^{2}+5x+x^{2}+2x+1=10\left(x+1\right)$$

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

$$2x^{2}+5x+2x+1=10\left(x+1\right)$$

Combine 5x and 2x to get 7x.

$$2x^{2}+7x+1=10\left(x+1\right)$$

Use the distributive property to multiply 10 by x+1.

$$2x^{2}+7x+1=10x+10$$

Subtract 10x from both sides.

$$2x^{2}+7x+1-10x=10$$

Combine 7x and -10x to get -3x.

$$2x^{2}-3x+1=10$$

Subtract 10 from both sides.

$$2x^{2}-3x+1-10=0$$

Subtract 10 from 1 to get -9.

$$2x^{2}-3x-9=0$$

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

$$a+b=-3 ab=2\left(-9\right)=-18$$

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 -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -3.

$$a=-6 b=3$$

Rewrite 2x^{2}-3x-9 as \left(2x^{2}-6x\right)+\left(3x-9\right).

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

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

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

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

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

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

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

Steps Using the Quadratic Formula

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

$$x^{2}+5x+\left(x+1\right)\left(x+1\right)=10\left(x+1\right)$$

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

$$x^{2}+5x+\left(x+1\right)^{2}=10\left(x+1\right)$$

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

$$x^{2}+5x+x^{2}+2x+1=10\left(x+1\right)$$

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

$$2x^{2}+5x+2x+1=10\left(x+1\right)$$

Combine 5x and 2x to get 7x.

$$2x^{2}+7x+1=10\left(x+1\right)$$

Use the distributive property to multiply 10 by x+1.

$$2x^{2}+7x+1=10x+10$$

Subtract 10x from both sides.

$$2x^{2}+7x+1-10x=10$$

Combine 7x and -10x to get -3x.

$$2x^{2}-3x+1=10$$

Subtract 10 from both sides.

$$2x^{2}-3x+1-10=0$$

Subtract 10 from 1 to get -9.

$$2x^{2}-3x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-9\right)}}{2\times 2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-9\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-9\right)}}{2\times 2}$$

Multiply -8 times -9.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2\times 2}$$

Add 9 to 72.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2\times 2}$$

Take the square root of 81.

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

The opposite of -3 is 3.

$$x=\frac{3±9}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±9}{4}$$

Now solve the equation x=\frac{3±9}{4} when ± is plus. Add 3 to 9.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{3±9}{4} when ± is minus. Subtract 9 from 3.

$$x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

$$x^{2}+5x+\left(x+1\right)\left(x+1\right)=10\left(x+1\right)$$

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

$$x^{2}+5x+\left(x+1\right)^{2}=10\left(x+1\right)$$

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

$$x^{2}+5x+x^{2}+2x+1=10\left(x+1\right)$$

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

$$2x^{2}+5x+2x+1=10\left(x+1\right)$$

Combine 5x and 2x to get 7x.

$$2x^{2}+7x+1=10\left(x+1\right)$$

Use the distributive property to multiply 10 by x+1.

$$2x^{2}+7x+1=10x+10$$

Subtract 10x from both sides.

$$2x^{2}+7x+1-10x=10$$

Combine 7x and -10x to get -3x.

$$2x^{2}-3x+1=10$$

Subtract 1 from both sides.

$$2x^{2}-3x=10-1$$

Subtract 1 from 10 to get 9.

$$2x^{2}-3x=9$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

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

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{9}{2}+\frac{9}{16}$$

Add \frac{9}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{81}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{9}{4} x-\frac{3}{4}=-\frac{9}{4}$$

Add \frac{3}{4} to both sides of the equation.

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