Question

a table for y=2x^2-8 is given solve each equation. a. 2x^2-8=0,b. 2x^2-8<0,c. 2x^2-8>0.

Answer 1

where is the table?

Answer 2

it is in the question

Answer 3

2x^2-8=0
2x^2=8
x^2=4
x=-2 or 2

Answer 4

got it?

Answer 5

is that forthe equation a.

Answer 6

yes

Answer 7

what about b and c

Answer 8

Did u understand it?

Answer 9

yes for thef irst one I did

Answer 10

2x^2-8<0
2x^2<8
x^2<4
x^2<(+-2)^2
Now, IF x^2<(+-a)^2 then -a<x<a

Answer 11

SO, can u complete it?

Answer 12

no do not get that one

Answer 13

2x^2-8<0
2x^2<8
x^2<4
x^2<(+-2)^2
IS it OKAY up to here?

Answer 14

yes

Answer 15

would it be 4

Answer 16

IF x^2<(+-a)^2 then -a<x<a

Answer 17

try to use it

Answer 18

I do notg et it

Answer 19

x^2<(+-2)^2
-2<x<2

Answer 20

got it?

Answer 21

yes but how do i figure it out

Answer 22

HINT: IF x^2<(+-a)^2
then -a<x<a

Answer 23

It is like formula

Answer 24

howdo i figurethef o rmula out

Answer 25

U want me to derive it?

Answer 26

what does that mean

Answer 27

hello what doyou m ean

Answer 28

I am sorry but I guess u need to revise your algebra once

Answer 29

soyoucannoth elp meany further

Answer 30

Sorry not so good at explaining things

Answer 31

ok Iwil l try someelse

Answer 32

@mathslover can help u?

Answer 33

He is good at explaining

Answer 34

I am well at explanation :)

Answer 35

socanyouhelpme mathsolver

Answer 36

I will be right back ... wait for 2 minutes please

Answer 37

\[
2x^2-8<0 \\
2x^2<8 \\
x^2<4 \\
\sqrt{x^2}<\sqrt{4}
\]Now it is important to remember that \(\sqrt{\quad}\) mean the "positive square root" (not the negative one) and that \(\sqrt{a^2}\) is the definition of of the absolute value.\[
|x| < 2
\]

Answer 38

To solve absolute value equations, we have to split it up into two cases.
Case 1: Assume \(x\) (the thing in the absolute value) is positive. \[
|x| < 2 \\
x<2
\] That was easy.
Case 2: Assume \(x\) is negative: \[
|x| < 2 \\
-x < 2 \\
x > -2
\]Remember that if \(x\) is negative, then \(|x| =-x\) because the absolute value bars had to flip the sign to make \(x\) positive.

Also remember that when you multiply/divide both sides of an inequality by a negative number, the equality flips. Hence \(<\) became \(>\) in this instance.

Answer 39

Notation allows us to conveniently combine the expressions \(x < 2\) and \(x > -2\) into \(-2 < x < 2\).

Does this help, @cdelomas ?

Answer 40

One thing worthy of committing to memory is that: \[
|x| < a \implies -a < x < a \\
|x| > a \implies x < -a,\quad x > a
\]

Answer 41

so for equation b is it squareroot x2>squareroot 4.