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