How do you solve $m^{2} + a \pm 16 = 0$ $m^2+a+-16=0$ ?

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How do you solve $m^{2} + a \pm 16 = 0$ $m^2+a+-16=0$ ?

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Answer 1 · 2016-04-07T11:42:24

$m = \pm \sqrt{- 16 - a}$ $m=+-sqrt(-16-a)$ (two solutions if $a \leq - 16$ $a<=-16$ ; none otherwise)
or
$m = \pm \sqrt{16 - a}$ $m=+-sqrt(16-a)$ (two solutions if $a < 16$ $a<16$ ; none otherwise)

Explanation:

$m^{2} + a \pm 16 = 0$ $m^2+a+-16=0$

$\to m^{2} = \pm 16 - a$ $rarr m^2= +-16-a$

$\to m = \pm \sqrt{\pm 16 - a}$ $rarr m=+-sqrt(+-16-a)$

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How do you solve $m^{2} + a \pm 16 = 0$ $m^2+a+-16=0$ ?

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