How do you solve t^2+16=0 using the quadratic formula?

Mar 2, 2018

$t = 4 i$
$t = - 4 i$

Explanation:

$a {x}^{2} + b x + c = 0$

where $a \ne 0$, but $b$ and $c$ can equal zero.

Here, $b = 0$. The expression can be rewritten as:

$1 {t}^{2} + 0 t + 16 = 0$

Now, we can input the values into the quadratic formula:

$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$t = \frac{- 0 \pm \sqrt{{0}^{2} - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

$t = \frac{\pm \sqrt{- 64}}{2}$

$t = \frac{\pm 8 i}{2}$

where $i = \sqrt{-} 1$

now, $t$ has $2$ possible solutions:

1) $t = \frac{+ 8 i}{2}$

$t = 4 i$

2) $t = \frac{- 8 i}{2}$

$= - 4 i$

The roots are imaginary, and so are the answers.

Another thing to be noted is that according to the Conjugate Pairs Theorem, if $\left(a + b i\right)$ is the root of a polynomial, $\left(a - b i\right)$ is also a root, and vice versa. The above fits this rule.