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

1 Answer
Jul 5, 2015

Answer:

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#