Question

How do you solve `x^2 + x +10 = 0` using the quadratic formula?

Answer 1

Use the quadratic formula to find roots:

`x =-1/2+-sqrt(39)/2 i`

Explanation:

`x^2+x+10 = 0` is of the form `ax^2+bx+c = 0` with `a=1`, `b=1` and `c=10`.

It has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = 1^2-(4*1*10) = 1-40 = -39`

Since this is negative, this quadratic equation has no Real roots.

It has a Complex conjugate pair of roots given by the quadratic formula:

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

`=(-b+-sqrt(Delta))/(2a)`

`=(-1+-sqrt(-39))/2`

`=(-1+-sqrt(39)i)/2`

`=-1/2+-sqrt(39)/2 i`

Answer 2

zero

Explanation:

`x^2+x+10=0` is in the form of `ax^2+bx+c=0`. Then you apply the quadratic formula

You type into the calculator this:

x=-1+√`1^2`- 4110/2*1

This equals to zero because a surd cannot be a negative number. (Some equations cannot be solved and this is one of them)

Answer 3

`x=-1/2+-(sqrt39)/2i`

Explanation:

`color(blue)(x^2+x+10=0`

This is a Quadratic equation (in form `ax^2+bx+c=0`)

Use Quadratic formula

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

Remember that `a,band c` are the coefficients of `x^2,xand10`

Where,

`color(red)(a=1,b=1,c=10`

And don't be afraid with the formula!

`rarrx=(-1+-sqrt(1^2-4(1)(10)))/(2(1))`

`rarrx=(-1+-sqrt(1-4(10)))/(2)`

`rarrx=(-1+-sqrt(1-40))/(2)`

`rarrx=(-1+-sqrt(-39))/(2)`

Oh! we cannot find the square root of `-39` because it is a negative number!.Don't worry,in such cases it is called a complex number
(in form `sqrt(-1)`) .It is represented as `i`

So,

`rarrx=(-1+-sqrt(39*-1))/(2)`

`rarrx=(-1+-sqrt39)/(2)i`

`color(green)(rArrx=(-1)/(2)+-(sqrt39)/(2)i`