How do you solve $- 2 x^{2} - 80 = 0$ $-2x^2-80=0$ ?

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Answer 1 · 2016-11-08T19:35:16

$x = \pm 2 \sqrt{10} i$ $x=+-2sqrt(10)i$

$- 2 x^{2} - 80 = 0$ $-2x^2-80=0$

$\to x^{2} + 40 = 0$ $rarr x^2+40=0$

$\to x^{2} = - 40 = i^{2} \cdot 2^{2} \cdot 10$ $rarr x^2=-40= i^2 * 2^2 * 10$

$\to x = \pm i 2 \sqrt{10} or \pm 2 \sqrt{10} i$ $rarr x = +- i2sqrt(10) or +-2sqrt(10)i$

Answer 2 · 2016-11-08T19:40:11

$- 2 x^{2} - 80 = 0$ $-2x^2 -80 =0$

Usually you would want a quadratic equation to be equal to 0, but in this case there is no $x$ $x$ term, so we will solve it by another method.

Re-arrange to give:

$- 2 x^{2} = 80 \leftarrow \div - 2$ $-2x^2 = 80" "larr div -2$

$x^{2} = - 40$ $x^2 = -40$

$x = \pm \sqrt{- 40}$ $x = +-sqrt(-40)$

There are no real solutions to this equation.

How do you solve -2x^2-80=0−2x2−80=0-2x^2-80=0?