Wotzdorfg

2021-09-11

Find the solutions of the equation.

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

lobeflepnoumni

Skilled2021-09-12Added 99 answers

Step 1

The given equation is:

${x}^{2}+4x+13=0$ ...(1)

As, the given equation is a quadratic equation.

The general form f a quadratic equation is:

$a{x}^{2}+bx+c=0$ ...(2)

So, finding the roots of any quadratic equation, we use:

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

On comparing equation (1) with the equation(2), we have:

a=1

b=4

c=13

Step 2

Solving it, we get:

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

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

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

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

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

Since,$\sqrt{-1}=i$

So,$x=-2\pm 3i$

Hence, the quadratic equation have complex roots.

Therefore, we have:

x=-2+3i

x=-2-3i

The required solution of the given quadratic equation.

The given equation is:

As, the given equation is a quadratic equation.

The general form f a quadratic equation is:

So, finding the roots of any quadratic equation, we use:

On comparing equation (1) with the equation(2), we have:

a=1

b=4

c=13

Step 2

Solving it, we get:

Since,

So,

Hence, the quadratic equation have complex roots.

Therefore, we have:

x=-2+3i

x=-2-3i

The required solution of the given quadratic equation.

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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