Question

How do you solve `x^2 + 5x + 7 = 0` using the quadratic formula?

Answer 1

`(-5+isqrt(3))/2` and `(-5-isqrt(3))/2`

Explanation:

For quadratic equations of the form:

`ax^2+bx+c`

The quadratic formula is given by:

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

From given equation we have:

`bba =1`

`bb(b)=5`

`bbc=7`

Putting these values in the quadratic formula:

`(-(5)+-sqrt((5)^2-4(1)(7)))/(2(1))=(-5+-sqrt(25-(28)))/(2)`

`=(-5+-sqrt(-3))/2`

We can write this in the following way:

`sqrt(-3)=sqrt(3xx-1)=sqrt(3)*sqrt(-1)`

If `sqrt(-1)=i`

Then:

`(-5+isqrt(3))/2` and `(-5-isqrt(3))/2`

These are known as complex roots.

Answer 2

`x=(-5+isqrt3)/2` or `(-5-isqrt3)/2`

Explanation:

Accordng to quadratic formula, solution of quadratic equation `ax^2+bx+c=0` is

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

Hence solution of `x^2+5x+7=0` is

`x=(-5+-sqrt(5^2-4*1*7))/2`

= `(-5+-sqrt(25-28))/2`

= `(-5+-sqrt(-3))/2`

i.e. `x=(-5+isqrt3)/2` or `(-5-isqrt3)/2`

Answer 3

No real solutions

Explanation:

`x^2+5x+7=0`

Use the quadratic formula with `a=1, b=5, c=7`

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

`x=(-5+-sqrt((5)^2-4(1)(7)))/((2)(1))`

`x=(-5+-sqrt(25 - 28))/2`

`x=(-5+-sqrt(-3))/2`

No real solution