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March 3, 2015

Posted by **Lisa D** on Sunday, June 27, 2010 at 9:04pm.

A) 2

B) 1

C) 0

Not sure how to work this problem

- Discriminant -
**bobpursley**, Sunday, June 27, 2010 at 9:25pmrearrange it.

7a^2+7a+2=0

discriminate= 49-4*7*2= negative number, no real solutions.

- Discriminant -
**MathMate**, Sunday, June 27, 2010 at 9:28pmFirst 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.

- Discriminant -
**drwls**, Sunday, June 27, 2010 at 9:30pmRewrite 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 10:17pmThanks 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 11:46pmOk, I got it. Sorry been having problems w/ this site.Thanks I got more..

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