Solve x+51/x=-6

$$x+ 5 \frac { 1 } { x } =-6$$

Solve for x

$$x=-5$$

$$x=-1$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$xx+5\times 1=-6x$$

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

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

Multiply 5 and 1 to get 5.

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

Add 6x to both sides.

$$x^{2}+5+6x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+6x+5=0$$

To solve the equation, factor x^{2}+6x+5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=6 ab=5$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=5$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

To find equation solutions, solve x+1=0 and x+5=0.

$$x=-1 x=-5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$xx+5\times 1=-6x$$

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

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

Multiply 5 and 1 to get 5.

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

Add 6x to both sides.

$$x^{2}+5+6x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+6x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\times 5=5$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=5$$

Rewrite x^{2}+6x+5 as \left(x^{2}+x\right)+\left(5x+5\right).

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

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

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

Factor out common term x+1 by using distributive property.

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

To find equation solutions, solve x+1=0 and x+5=0.

$$x=-1 x=-5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$xx+5\times 1=-6x$$

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

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

Multiply 5 and 1 to get 5.

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

Add 6x to both sides.

$$x^{2}+5+6x=0$$

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.

$$x^{2}+6x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-6±\sqrt{36-20}}{2}$$

Add 36 to -20.

$$x=\frac{-6±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-6±4}{2}$$

Now solve the equation x=\frac{-6±4}{2} when ± is plus. Add -6 to 4.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-6±4}{2} when ± is minus. Subtract 4 from -6.

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

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=-1 x=-5$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$xx+5\times 1=-6x$$

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

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

Multiply 5 and 1 to get 5.

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

Add 6x to both sides.

$$x^{2}+5+6x=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

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

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-5+3^{2}$$

Square 3.

$$x^{2}+6x+9=-5+9$$

Add -5 to 9.

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

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+3=2 x+3=-2$$

Subtract 3 from both sides of the equation.

$$x=-1 x=-5$$