Question

How do you find the zeroes of `f(x) = x^2 + 10x+9` using the quadratic formula?

Answer 1

The zeroes are at `x = -1` and `x = -9`.

Here's how I did it:

Explanation:

This equation is in the form of `ax^2 + bx + c = y`, or standard form. In standard form, we can use quadratic formula to solve for the zeroes. The quadratic formula is `x = (-b +- sqrt(b^2 - 4ac))/(2a)`.

In your question `f(x) = x^2 + 10x + 9`:
`a = 1`
`b = 10`
`c = 9`

So now we can plug in these values in the quadratic formula:
`x = (-10 +- sqrt(10^2 - 4(1)(9)))/(2(1))`

...And simplify
`x = (-10 +- sqrt(100 - 36))/2`

`x = (-10 +- sqrt(64))/2`

`x = (-10 +- 8)/2`

So now this splits into two equations:
`x = (-10+8)/2` and `x = (-10-8)/2`.

`x = -2/2` and `x = -18/2`

`x = -1` and `x = -9`.

Hope this helps!