Posted by Lisa D on Sunday, June 27, 2010 at 9:04pm.
Use the discriminant to determine how many realnumber solutions the equation has 1  7a2 = 7a  2
A) 2
B) 1
C) 0
Not sure how to work this problem

Discriminant  bobpursley, Sunday, June 27, 2010 at 9:25pm
rearrange it.
7a^2+7a+2=0
discriminate= 494*7*2= negative number, no real solutions. 
Discriminant  MathMate, Sunday, June 27, 2010 at 9:28pm
First arrange the terms in the standard form of a quadratic:
1  7a2 = 7a  2
7a²7a3=0
The variable is a, and the coefficients of the a², a and constant terms are respectively
A=7, B=7, C=3
The discriminant is the expressino
D=(B²4AC)
and the roots of the equation are:
X1,X2= (B±√(B²4AC))/2A
Therefore
if D>0, there are two roots, if D=0, there is one root (actually two coincident roots), and if D<0, the roots are complex, or generally considered "no root".
Post your answer for checking if you wish. 
Discriminant  drwls, Sunday, June 27, 2010 at 9:30pm
Rewrite your equation in standard form:
7a^2 7a 3 = 0
They are trying to confuse you by using a as a variable. Think of your equation as
7x^2 7x 3 = 0
It does not matter what symbol you giove to the unknown variable; just don't call it "a".
The numbers a, b, and c in the equation are 7, 7 and 3.
The discriminant is b^2 4ac = 49+84 = 133
Since it is positive, there are two real solutions.
They are
x = (1/14)(7 +/ sqrt(133)]= 1.324 and
0.324 
Discriminant  Lisa D, Sunday, June 27, 2010 at 10:17pm
Thanks for the Explaning this problem drwls. Ok, the first time I worked the problem I got 0 then I reworked it and got 1

Discriminant  drwls, Sunday, June 27, 2010 at 11:32pm
133 should be 133 in my previous answer.
There is NOT one root. There are two, if I did the numbers right. 
Discriminant  Lisa D, Sunday, June 27, 2010 at 11:46pm
Ok, I got it. Sorry been having problems w/ this site.Thanks I got more..