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

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Answer 1 · 2015-07-05T13:47:19

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

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 $9n^2-3n-8=-10$

How to use the discriminant to find out how many real number roots an equation has for 9n^2 - 3n - 8= -109n^2 - 3n - 8= -10?