How do you solve $x^{2} + 4 = 0$ $x^2+ 4 = 0$ ?

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How do you solve $x^{2} + 4 = 0$ $x^2+ 4 = 0$ ?

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Answer 1 · 2018-04-14T04:01:17

$x = \pm 2 i$ $x=+-2i " "$

Explanation:

$x^{2} + 4 = 0$ $x^2+4=0 " "$ subtract 4 on both sides to get

$x^{2} = - 4$ $x^2 = -4 " "$ Take the square root of both sides to get

$x = \pm 2 i$ $x=+-2i " "$

Answer 2 · 2018-04-14T04:08:03

$x = \pm 2 i$ $x=+-2i$

Explanation:

For this, you need a concept of imaginary numbers

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

$\Rightarrow x^{2} + 4 - 4 = 0 - 4$ $rArr x^2+4-4=0-4$

(Subtracting $4$ $4$ from both sides)

$\Rightarrow x^{2} = - 4$ $rArr x^2=-4$

$\Rightarrow x = \sqrt{- 4}$ $rArr x=sqrt(-4)$

$\Rightarrow x = \sqrt{- 1 \times 4}$ $rArr x=sqrt(-1xx4)$

$\Rightarrow x = \pm 2 \sqrt{- 1}$ $rArr x=+-2sqrt(-1)$

(The value of $\sqrt{- 1}$ $sqrt(-1)$ is $i$ $i$ )

$\Rightarrow x = \pm 2 i$ $rArr x=+-2i$

Hope this helps :)

Answer 3 · 2018-04-14T04:11:43

No real solution

Explanation:

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

Start by subtracting $4$ $4$ from both sides

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

$x^{2} = - 4$ $x^2 = -4$

Now we can take square root

$x = \pm \sqrt{- 4}$ $x = +-sqrt(-4)$

Thus,

No real solutions

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How do you solve $x^{2} + 4 = 0$ $x^2+ 4 = 0$ ?

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