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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.

Mathematics
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where is the table?
it is in the question
2x^2-8=0 2x^2=8 x^2=4 x=-2 or 2

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Other answers:

got it?
is that forthe equation a.
yes
what about b and c
Did u understand it?
yes for thef irst one I did
2x^2-8<0 2x^2<8 x^2<4 x^2<(+-2)^2 Now, IF x^2<(+-a)^2 then -a
SO, can u complete it?
no do not get that one
2x^2-8<0 2x^2<8 x^2<4 x^2<(+-2)^2 IS it OKAY up to here?
yes
would it be 4
IF x^2<(+-a)^2 then -a
try to use it
I do notg et it
x^2<(+-2)^2 -2
got it?
yes but how do i figure it out
HINT: IF x^2<(+-a)^2 then -a
It is like formula
howdo i figurethef o rmula out
U want me to derive it?
what does that mean
hello what doyou m ean
I am sorry but I guess u need to revise your algebra once
soyoucannoth elp meany further
Sorry not so good at explaining things
ok Iwil l try someelse
@mathslover can help u?
He is good at explaining
I am well at explanation :)
socanyouhelpme mathsolver
I will be right back ... wait for 2 minutes please
\[ 2x^2-8<0 \\ 2x^2<8 \\ x^2<4 \\ \sqrt{x^2}<\sqrt{4} \]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.\[ |x| < 2 \]
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.
Notation allows us to conveniently combine the expressions \(x < 2\) and \(x > -2\) into \(-2 < x < 2\). Does this help, @cdelomas ?
One thing worthy of committing to memory is that: \[ |x| < a \implies -a < x < a \\ |x| > a \implies x < -a,\quad x > a \]
so for equation b is it squareroot x2>squareroot 4.

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