Question

I REALLY NEED HELP!!
what is the simplified form of the expression sqrt 16c^4- sqrt c^2+3 sqrt c^2+ sqrt 9c^3

Answer 1

Is this the problem?
\( \sqrt {16c^4} - \sqrt{ c^2}+3 \sqrt {c^2}+ \sqrt {9c^3}\)

Answer 2

no the last sqrt is
sqrt 9c^2

Answer 3

thank you

Answer 4

\(\sqrt {16c^4} - \sqrt{ c^2}+3 \sqrt {c^2}+ \sqrt {9c^2}\)

In general:
\( \sqrt{a^2} = |a| \)
\( \sqrt{a^4} = a^2 \)

Answer 5

yes that is right! so do u think you can help me with it?

Answer 6

im taking a module test on FLVS and I seem to be having a hard time with it

Answer 7

Have you been told that c is positive or nonnegative?

Answer 8

um.no?

Answer 9

Ok, then you need to use the absolute value.

Every time you see sqrt(c^2), you can replace it by |c|.
When you see sqrt(c^4), you can replace it by c^2.

Answer 10

\(\sqrt {16c^4} - \sqrt{ c^2}+3 \sqrt {c^2}+ \sqrt {9c^2}\)

\(= 4c^2 - |c| + 3|c| + |3c| \)

\(= 4c^2 - |c| + 3|c| + 3|c| \)

Terms with |c| are like terms and can be combined together.

Answer 11

ok..but I am still very confused on how to solve this problem

Answer 12

sorry

Answer 13

\(= 4c^2 +5 |c| \)

Answer 14

ohh! I see how you kinda did the subsition method ;) thank you for helping me with this! I couldn't have done with out you! line no joke!

Answer 15

You're welcome.

Answer 16

If I need help again.. (I don't right now) would you be willing to help me?

Answer 17

im sorry but could you help me again on this:
which factor would you cancel from the numerator and denominator to simplify x^2-3x-4 over x^2-16

Answer 18

\( \dfrac{x^2 - 3x - 4}{x^2 - 16} \)

You need to factor the numerator and the denominator.

\( \dfrac{(x - 4)(x + 1)}{(x + 4)(x - 4) } \)

Now you can divide the numerator and denominator by the same factor, x - 4.