Question

What are the roots of the equation `x^2+4x-16=0`?

Answer 1

`x=-2+-2sqrt(5)`

Explanation:

This quadratic equation is in the form `ax^2+bx+c`, where `a=1`, `b=4`, and `c=-16`. In order to find the roots, we can use the quadratic formula below.

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

`x=(-4+-sqrt(4^2-4(1)(-16)))/(2(1))`

`x=(-4+-sqrt(80))/(2)`

`x=(-4+-4sqrt(5))/(2)`

`x=-2+-2sqrt(5)`

Answer 2

See a solution process below:

Explanation:

We can use the quadratic formula to find the roots for this equation. The quadratic formula states:

For `ax^2 + bx + c = 0`, the values of `x` which are the solutions to the equation are given by:

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

Substituting `1` for `a`; `4` for `b` and `-16` for `c` gives:

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

`x = (-4 +- sqrt(16 - (-64)))/2`

`x = (-4 +- sqrt(80))/2`

`x = (-4 + sqrt(16 * 5))/2` and `x = (-4 - sqrt(16 * 5))/2`

`x = (-4 + (sqrt(16)sqrt(5)))/2` and `x = (-4 - (sqrt(16)sqrt(5)))/2`

`x = (-4 + 4sqrt(5))/2` and `x = (-4 - 4sqrt(5))/2`

`x = -2 + 2sqrt(5)` and `x = -2 - 2sqrt(5)`