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cdelomas
Group Title
a table for y=2x^28 is given solve each equation. a. 2x^28=0,b. 2x^28<0,c. 2x^28>0.
 one year ago
 one year ago
cdelomas Group Title
a table for y=2x^28 is given solve each equation. a. 2x^28=0,b. 2x^28<0,c. 2x^28>0.
 one year ago
 one year ago

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sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
where is the table?
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
it is in the question
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
2x^28=0 2x^2=8 x^2=4 x=2 or 2
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
got it?
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
is that forthe equation a.
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
what about b and c
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Did u understand it?
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
yes for thef irst one I did
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
2x^28<0 2x^2<8 x^2<4 x^2<(+2)^2 Now, IF x^2<(+a)^2 then a<x<a
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
SO, can u complete it?
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
no do not get that one
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
2x^28<0 2x^2<8 x^2<4 x^2<(+2)^2 IS it OKAY up to here?
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
would it be 4
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
IF x^2<(+a)^2 then a<x<a
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
try to use it
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
I do notg et it
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
x^2<(+2)^2 2<x<2
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
got it?
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
yes but how do i figure it out
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
HINT: IF x^2<(+a)^2 then a<x<a
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
It is like formula
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
howdo i figurethef o rmula out
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
U want me to derive it?
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
what does that mean
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
hello what doyou m ean
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
I am sorry but I guess u need to revise your algebra once
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
soyoucannoth elp meany further
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Sorry not so good at explaining things
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
ok Iwil l try someelse
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
@mathslover can help u?
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
He is good at explaining
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
I am well at explanation :)
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
socanyouhelpme mathsolver
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
I will be right back ... wait for 2 minutes please
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.0
\[ 2x^28<0 \\ 2x^2<8 \\ x^2<4 \\ \sqrt{x^2}<\sqrt{4} \]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.\[ x < 2 \]
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.0
Notation allows us to conveniently combine the expressions \(x < 2\) and \(x > 2\) into \(2 < x < 2\). Does this help, @cdelomas ?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.0
One thing worthy of committing to memory is that: \[ x < a \implies a < x < a \\ x > a \implies x < a,\quad x > a \]
 one year ago

cdelomas Group TitleBest ResponseYou've already chosen the best response.0
so for equation b is it squareroot x2>squareroot 4.
 one year ago
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