Question

How to use the discriminant to find out how many real number roots an equation has for `9n^2 - 3n - 8= -10`?

Answer 1

There is no real number root to `9n^2-3n-8=-10`

Explanation:

The first step is to change the equation to the form:

`an^2+bn+c=0`

To do so, you must do:

`9n^2-3n-8+10=-cancel(10)+cancel10`
`rarr 9n^2-3n+2=0`

Then, you must calculate the discriminant:

`Delta=b^2-4*a*c`

In your case:
`a=9`
`b=-3`
`c=2`

Therefore:

`Delta=(-3)^2-4*9*2=9-72=-63`

Depending on the result, you can conclude how many real solutions exist:

if `Delta>0`, there are two real solutions:
`rarr n_+=(-b+sqrtDelta)/(2a)` and `n_(-)=(-b-sqrtDelta)/(2a)`

if `Delta=0`, there is one real solution:
`rarr n_0=(-b)/(2a)`

if `Delta<0`, there is no real solution.

In your case, `Delta=-63<0`, therefore there is no real number root to `9n^2-3n-8=-10`