Question

Use the discriminant to determine how many real-number solutions the equation has 1 - 7a2 = -7a - 2

A) 2
B) 1
C) 0
Not sure how to work this problem

Answer 1

rearrange it.

7a^2+7a+2=0

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

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

-133 should be 133 in my previous answer.
There is NOT one root. There are two, if I did the numbers right.

Answer 6

Ok, I got it. Sorry been having problems w/ this site.Thanks I got more..