Question

Help with this please?

Answer 1

I've gotten the inequality to equal 0 so I had -2x^2 - 4x - 1 = 0, since it can't be factored I'm using the quadratic formula (-b +/- sqrtb^2 - 4ac over 2a) to the point where I've gotten to 4 +/- sqrt 24 over 4..I'm not sure what to do next?

Answer 2

I see... you've gotten to the roots, then.
There are only three intervals to worry about, they're the ones defined by your roots.

Answer 3

Your inequality was \(\large 2x^2 +4x + 1 > 0\) yes?

Answer 4

no my original eq2uality was 4x > -1 - 2x^

Answer 5

2x^2*

Answer 6

I know that, but this is equivalent, right?
I just brought the terms over to the left side.

Answer 7

I think so Yea

Answer 8

And you factored it, you got the roots.
So you want that thing on the left to be positive (greater than zero)
So what I want you to do is consider these three intervals:
\[\large \left(-\infty, \frac{4-\sqrt{24}}{4}\right)\quad,\quad\left(\frac{4-\sqrt{24}}{4},\frac{4+\sqrt{24}}{4}\right)\quad,\quad \left(\frac{4+\sqrt{24}}{4},\infty\right)\]

Answer 9

Kept them open since obviously, the roots themselves won't be solutions, since we want the left-polynomial to be strictly positive.

Answer 10

I don't really understand what we're doing haha, this isn't in my book so I'm kinda lost xD

Answer 11

Sorry... mind telling me what the book says?

Answer 12

sure one second

Answer 13

hmm, your book is Wrong

Answer 14

It is?

Answer 15

yes, the answer provided is does not answer the question,

Answer 16

oh there's another page I didn't include

Answer 17

oh

Answer 18

In order to solve my problem, should I sqrt 24 and round it to the nearest tenth and go from there?

Answer 19

\[4x>-1-2x^2\\2x^2+4x+1>0\]
the related equation
\[2x^2+4x+1=0\]
\[x=\frac{-4\pm\sqrt{4^2-4(2)(1)}}{2(2)}\\=\frac{-4\pm\sqrt{16-8}}{4}\\=\]

Answer 20

ohh you added it to the other side, I subtracted 4x okay

Answer 21

16- 8 is 8..so we're still stuck with a number that can't be sqrted

Answer 22

well 8 = 4 . 2 so √8 = √4 . √2 = 2√2

Answer 23

so now we have -4 +/- 2sqrt2 over 4 so now we have to do the +/- part?

Answer 24

so you have
x=(-4 ± 2√2) /4


which means there are two solutions

x_1=(-4 + 2√2) /4

x_2=(-4 - 2√2) /4


solve each case

Answer 25

x_1=(-4 + 2√2) /4

=-4/4 + 2√2/4

Answer 26

so that equals -1 + sqrt2?

Answer 27

yep

what about x_2

Answer 28

-1 - sqrt2?

Answer 29

yes

Answer 30

so those are the two boundary points

Answer 31

x=x_1
and
x=x_2

Answer 32

√2 is about one and a half units

Answer 33

so about 1.4?

Answer 34

yeah,

Answer 35

hang on, wait a minute, what happen on those boundary points

x_1=(-4 + 2√2) /4

=-4/4 + 2√2/4

=-1 - (2/4) √2
=

Answer 36

oh so it'd be -1 - .5sqrt2 which is .5 + sqrt2?

Answer 37

=-1 - (2/4) √2
=-1 - (1/2) √2
=-1 -(√2)/2
=-1 -.5√2

Answer 38

so -1.5 + sqrt2?

Answer 39

you cant simply further

Answer 40

ohh right, its connected to the sqrt2

Answer 41

can you find the points on the number line ?

Answer 42

I think so...-.5 * sqrt2 is -.7 so -1.7?

Answer 43

yeah that is right (to 1 decimal place)

Answer 44

what about the other boundary value? (again to 1 d.p.)

Answer 45

-2.4?

Answer 46

=-1 +.5√2

Answer 47

-.3

Answer 48

yes, now can you mark -1.7 and -.3 on the number line ?

Answer 49

Yes