Get the solution to "How do you solve x2+16=0 using the quadratic..."

High School
Algebra
Algebra I
Quadratic function and equation

Lennie Davis

Answered question

2021-12-20

How do you solve

x^{2} + 16 = 0

using the quadratic formula?

Answer & Explanation

Jeffery Autrey

Beginner2021-12-21Added 35 answers

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

then

x = \frac{1}{2 a} (- b \pm \sqrt{b^{2} - 4 a c}

For

x^{2} + 16 = 0

a = 1, b = 0, c = 0

Substituting

x = \frac{1}{2 \times 1} (- 0 \pm \sqrt{0^{2} - 4 \times 1 \times 16})

= \frac{1}{2} \times \pm \sqrt{- 64}

= \pm \sqrt{- 16}

godsrvnt0706

Beginner2021-12-22Added 31 answers

The quadratic formula is given by:

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

The equation in standard form is

x^{2} + 0 x + 16

x^{2} = 0 x + 16

and

a = 1, b = 0, c = 16

Plug in:

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

\frac{0 \pm \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{\pm \sqrt{64 \cdot - 1}}{2}

\frac{\pm \sqrt{64} \cdot \sqrt{- 1}}{2}

\frac{\pm 8 i}{2}

\pm 4 i

RizerMix

Expert2021-12-29Added 656 answers

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

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