Solved on Feb 06, 2024

Find all complex solutions to the equation x+2+5x=0x + 2 + \frac{5}{x} = 0.

STEP 1

Assumptions
1. We are solving the equation x+2+5x=0x + 2 + \frac{5}{x} = 0.
2. Solutions may be real or complex numbers.

STEP 2

To solve the equation, we need to find a common denominator for the terms involving xx. In this case, the common denominator is xx.
x(x+2)+5=0x(x + 2) + 5 = 0

STEP 3

Now, distribute the xx across the terms in the parentheses.
x2+2x+5=0x^2 + 2x + 5 = 0

STEP 4

We have a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0. We can solve for xx using the quadratic formula:
x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

STEP 5

Identify the coefficients aa, bb, and cc from the quadratic equation x2+2x+5=0x^2 + 2x + 5 = 0.
a=1,b=2,c=5a = 1, b = 2, c = 5

STEP 6

Substitute the coefficients into the quadratic formula.
x=2±2241521x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}

STEP 7

Simplify the expression inside the square root.
x=2±4202x = \frac{-2 \pm \sqrt{4 - 20}}{2}

STEP 8

Calculate the value inside the square root.
x=2±162x = \frac{-2 \pm \sqrt{-16}}{2}

STEP 9

Since the value inside the square root is negative, we will have complex solutions. Recall that 1=i\sqrt{-1} = i.
x=2±4i2x = \frac{-2 \pm 4i}{2}

STEP 10

Divide both terms in the numerator by the denominator.
x=1±2ix = -1 \pm 2i

STEP 11

Write the solutions as a list of complex numbers.
The solutions are 1+2i,12i-1 + 2i, -1 - 2i.

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