Solve x+4=5/x

$$x + 4 = \frac { 5 } { x }$$

Solve for x

$$x=-5$$

$$x=1$$

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.

$$xx+x\times 4=5$$

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

$$x^{2}+x\times 4=5$$

Subtract 5 from both sides.

$$x^{2}+x\times 4-5=0$$

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.

$$x^{2}+4x-5=0$$

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$x=\frac{-4±\sqrt{16+20}}{2}$$

Add 16 to 20.

$$x=\frac{-4±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-4±6}{2}$$

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

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

Divide 2 by 2.

$$x=1$$

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

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

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.

$$xx+x\times 4=5$$

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

$$x^{2}+x\times 4=5$$

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.

$$x^{2}+4x=5$$

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

$$x^{2}+4x+2^{2}=5+2^{2}$$

Square 2.

$$x^{2}+4x+4=5+4$$

Add 5 to 4.

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

Factor x^{2}+4x+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}.

$$\left(x+2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{9}$$

Simplify.

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

Subtract 2 from both sides of the equation.

$$x=1 x=-5$$