Lennie Davis

2021-12-20

How do you solve ${x}^{2}+16=0$ using the quadratic formula?

Jeffery Autrey

If $a{x}^{2}+bx+c=0$
then $x=\frac{1}{2a}\left(-b±\sqrt{{b}^{2}-4ac}$
For ${x}^{2}+16=0$

Substituting
$x=\frac{1}{2×1}\left(-0±\sqrt{{0}^{2}-4×1×16}\right)$
$=\frac{1}{2}×±\sqrt{-64}$
$=±\sqrt{-16}$

godsrvnt0706

The quadratic formula is given by:
$\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
The equation in standard form is
${x}^{2}+0x+16$
${x}^{2}=0x+16$
and $a=1,b=0,c=16$
Plug in:
$\frac{-0±\sqrt{{0}^{2}-4\cdot 1\cdot 16}}{2\cdot 1}$
$\frac{0±\sqrt{-64}}{2}$
The number under the square root is negative, so there are no real solutions.
Finding the imaginary solutions ($i=\sqrt{-1}$):
$\frac{±\sqrt{64\cdot -1}}{2}$
$\frac{±\sqrt{64}\cdot \sqrt{-1}}{2}$
$\frac{±8i}{2}$
$±4i$

RizerMix

An alternative approach
${x}^{2}+16=0$
$⇒{x}^{2}=-16$
this has no real solutioons but will have 2 complex solutions
take square root of both sides
$⇒x=±\sqrt{-16}$ note plus or minus
$⇒x=±\sqrt{16×1}=±4i$

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