Question

4x^2+1>-4x
I don't know how to solve the first interval

Answer 1

4x^2+1>-4x

4x^2+1+4x>-4x+4x

4x^2+4x+1 > 0

now you need to solve 4x^2+4x+1 = 0 for x

Answer 2

Bring everything onto one side of the inequality sign, then set it equal to 0. factor it and you're good!

Answer 3

okay I got (2x+1)(2x+1) but I'm guessing thats not right?

Answer 4

you should have (2x+1)(2x+1) = 0

so x = ??

Answer 5

-1/2

Answer 6

notice how (2x+1)(2x+1) is the same as (2x+1)^2

Answer 7

(2x+1)^2 is either positive or 0

it will never be negative

Answer 8

so (2x+1)^2 > 0 is true for all real numbers x

Answer 9

well actually excluding x = -1/2

Answer 10

I'm confused on what the answer is

Answer 11

The answer is

\[\Large \left(-\infty, -\frac{1}{2}\right)\cup\left(-\frac{1}{2},\infty\right)\]

because we're just excluding the value -1/2 from the solution set

Answer 12

it says its wrong, but i noticed i wrote the original equation wrong its a \[\ge \] instead of >

Answer 13

how does that change it?

Answer 14

oh, then the solution set is all real numbers

Answer 15

in interval notation that would be

\[\Large \left(-\infty,\infty\right)\]

Answer 16

thank you so much

Answer 17

you're welcome