Question

Let f(x)=3x+1 and g(x)=2-x. Determine values for a such that f(a^2)=g(2a). How do you solve this? Please help D:

Answer 1

in f(x) substute a^2 for x
in g(x) use 2a in place of x

you are checking to see if you get the same values on both sides

Answer 2

\[3a ^{2}+1=2-2a\]
\[3a ^{2}+2a-1=0\]
(a+1)(3a-1)=0
a=-1
a=1/3

Answer 3

Ok well, since you're trying to find where f(a^2) = f(2a), we have to figure out what f(a^2) is and what g(2a) is.

f(a^2) = 3(a^2) + 1

f(a^2) = 3a^2 + 1

now let's plug in 2a for g(x):

g(2a) = 2 - 2a

Ok since, you need to find when f(a^2) = g(2a) just set them equal to each other:

3a^2 + 1 = 2 -2a

3a^2 + 2a + 1 - 2 = 0

3a^2 + 2a - 1 = 0

If you factor that, you get:

(3a + 1) (a - 1) = 0

therefore a=(1/3), (1)

Answer 4

sorry i meant (3a - 1) (a+1)

which means it should be:

a = (1/3), (-1)

Answer 5

Did you follow b100?

Answer 6

Yes that's the answer in the textbook! Thanks everyone! :)

Answer 7

My post follows JuanitaM, do what she suggested, and then my post would be a continuation of the first post. Hermeezey post shows the complete solution starting with what was given in the problem. Just wondering if you follow?

Answer 8

O.K. great, good luck in your studies.