Question

What are the zeros of the quadratic function `f(x)=8x^2-16x-15`?

Answer 1

`x = (16+-sqrt(736))/16` or `x = (4+-sqrt(46))/4`

Explanation:

In order to solve this quadratic formula, we will use the quadratic formula, which is `(-b+-sqrt(b^2-4ac))/(2a)`.

In order to use it, we need to understand which letter means what. A typical quadratic function would look like this: `ax^2 + bx + c`. Using that as a guide, we will assign each letter with their corresponding number and we get `a=8`, `b=-16`, and `c=-15`.

Then it is a matter of plugging in our numbers into the quadratic formula. We will get: `(-(-16)+-sqrt((-16)^2-4(8)(-15)))/(2(8))`.

Next, we will cancel out signs and multiply, which we will then get:
`(16+-sqrt(256+480))/16`.

Then we will add the numbers in the square root and we get `(16+-sqrt(736))/16`.

Looking at `sqrt(736)` we can probably figure out that we can simplify it. Let's use `16`. Dividing `736` by `16`, we will get `46`. So the inside becomes `sqrt(16*46)`. `16` is a perfect square root and the square of it is `4`. So carrying out `4`, we get `4sqrt(46)`.

Then our previous answer, `(16+-sqrt(736))/16`, becomes `(16+-4sqrt(46))/16`.

Notice that `4` is a factor of `16`. So taking our `4` from the numerator and denominator: `(4/4)(4+-sqrt(46))/4`. The two fours cancel out and our final answer is:

`(4+-sqrt(46))/4`.