Solve x+5/x^2+6x+5=1

$$\frac { x + 5 } { x ^ { 2 } + 6 x + 5 } = 1$$

Solve for x

$$x=0$$

Steps Using Factoring

Variable x cannot be equal to any of the values -5,-1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(x+5\right).

$$x+5=\left(x+1\right)\left(x+5\right)$$

Use the distributive property to multiply x+1 by x+5 and combine like terms.

$$x+5=x^{2}+6x+5$$

Subtract x^{2} from both sides.

$$x+5-x^{2}=6x+5$$

Subtract 6x from both sides.

$$x+5-x^{2}-6x=5$$

Combine x and -6x to get -5x.

$$-5x+5-x^{2}=5$$

Subtract 5 from both sides.

$$-5x+5-x^{2}-5=0$$

Subtract 5 from 5 to get 0.

$$-5x-x^{2}=0$$

Factor out x.

$$x\left(-5-x\right)=0$$

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

$$x=0 x=-5$$

Variable x cannot be equal to -5.

$$x=0$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,-1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(x+5\right).

$$x+5=\left(x+1\right)\left(x+5\right)$$

Use the distributive property to multiply x+1 by x+5 and combine like terms.

$$x+5=x^{2}+6x+5$$

Subtract x^{2} from both sides.

$$x+5-x^{2}=6x+5$$

Subtract 6x from both sides.

$$x+5-x^{2}-6x=5$$

Combine x and -6x to get -5x.

$$-5x+5-x^{2}=5$$

Subtract 5 from both sides.

$$-5x+5-x^{2}-5=0$$

Subtract 5 from 5 to get 0.

$$-5x-x^{2}=0$$

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.

$$-x^{2}-5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2\left(-1\right)}$$

Take the square root of \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±5}{-2}$$

Now solve the equation x=\frac{5±5}{-2} when ± is plus. Add 5 to 5.

$$x=\frac{10}{-2}$$

Divide 10 by -2.

$$x=-5$$

Now solve the equation x=\frac{5±5}{-2} when ± is minus. Subtract 5 from 5.

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

The equation is now solved.

$$x=-5 x=0$$

Variable x cannot be equal to -5.

$$x=0$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,-1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(x+5\right).

$$x+5=\left(x+1\right)\left(x+5\right)$$

Use the distributive property to multiply x+1 by x+5 and combine like terms.

$$x+5=x^{2}+6x+5$$

Subtract x^{2} from both sides.

$$x+5-x^{2}=6x+5$$

Subtract 6x from both sides.

$$x+5-x^{2}-6x=5$$

Combine x and -6x to get -5x.

$$-5x+5-x^{2}=5$$

Subtract 5 from both sides.

$$-5x-x^{2}=5-5$$

Subtract 5 from 5 to get 0.

$$-5x-x^{2}=0$$

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.

$$-x^{2}-5x=0$$

Divide both sides by -1.

$$\frac{-x^{2}-5x}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{5}{-1}\right)x=\frac{0}{-1}$$

Divide -5 by -1.

$$x^{2}+5x=\frac{0}{-1}$$

Divide 0 by -1.

$$x^{2}+5x=0$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=\frac{25}{4}$$

Factor x^{2}+5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{5}{2} x+\frac{5}{2}=-\frac{5}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=0 x=-5$$

Variable x cannot be equal to -5.

$$x=0$$