Posted by Lisa D on Sunday, June 27, 2010 at 9:04pm.
rearrange it.
7a^2+7a+2=0
discriminate= 49-4*7*2= negative number, no real solutions.
First arrange the terms in the standard form of a quadratic:
1 - 7a2 = -7a - 2
7a²-7a-3=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=(B²-4AC)
and the roots of the equation are:
X1,X2= (-B±√(B²-4AC))/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.
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
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
-133 should be 133 in my previous answer.
There is NOT one root. There are two, if I did the numbers right.
Ok, I got it. Sorry been having problems w/ this site.Thanks I got more..
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