Question

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

Answer 1

If `x^2+16=0`

`x=+-sqrt(-16)`

Explanation:

If `ax^2+bx+c=0,`
then
`x=1/(2a)(-b+-sqrt(b^2-4ac))`

For `x^2+16=0`

`a=1, b=0, c=16`
Substituting

`x=1/(2xx1)(-0+-sqrt(0^2-4xx1xx16))`

`=1/2xx+-sqrt(-64)`

`=+-sqrt(-16)`

Answer 2

`+-4i`

Explanation:

The quadratic formula is given by:

`(-b+-sqrt(b^2-4ac))/(2a)`

The equation in standard form is

`x^2+0x+16`

and `a=1,b=0,c=16`

Plug in:

`(-0+-sqrt(0^2-4*1*16))/(2*1)`

`(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)`):

`(+-sqrt(64*-1))/2`

`(+-sqrt(64)*sqrt(-1))/2`

`(+-8i)/2`

`+-4i`

Answer 3

`x=+-4i`

Explanation:

`"An alternative approach"`

`x^2+16=0`

`rArrx^2=-16`

`"this has no real solutions but will have 2 complex"`
`"solutions"`

`color(blue)"take square root of both sides"`

`rArrx=+-sqrt(-16)larrcolor(blue)"note plus or minus"`

`rArrx=+-sqrt(16xx-1)=+-4i`

Answer 4

`x=(0±sqrt(−64))/2`

Explanation:

`x=(−b±sqrt(b^2−4ac))/(2a)`

`x=(−(0)±sqrt((0)^2−4(1)(16)))/(2(1))`

`x=(0±sqrt(−64))/2`