Expert answers to "Find the solutions of the equation. x2+4x+13=0"

Wotzdorfg

Answered question

2021-09-11

Find the solutions of the equation.

x^{2} + 4 x + 13 = 0

lobeflepnoumni

Skilled2021-09-12Added 99 answers

Step 1
The given equation is:

x^{2} + 4 x + 13 = 0

...(1)
As, the given equation is a quadratic equation.
The general form f a quadratic equation is:

a x^{2} + b x + c = 0

...(2)
So, finding the roots of any quadratic equation, we use:

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

On comparing equation (1) with the equation(2), we have:
a=1
b=4
c=13
Step 2
Solving it, we get:

x = \frac{- 4 \pm \sqrt{{(- 4)}^{2} - 4 (1) (13)}}{2 (1)}

x = \frac{- 4 \pm \sqrt{16 - 52}}{2}

x = \frac{- 4 \pm \sqrt{- 36}}{2}

x = - 2 \pm \sqrt{- 9}

x = - 2 \pm 3 \sqrt{- 1}

Since,

\sqrt{- 1} = i

So,

x = - 2 \pm 3 i

Hence, the quadratic equation have complex roots.
Therefore, we have:
x=-2+3i
x=-2-3i
The required solution of the given quadratic equation.

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