Solve for x. 8x^2-5=11
All right! Do you know what the first step to solving the equation would be at all?
nope !:(
Basically, what you're going to do is try isolate x (get it by itself) in order to solve for it.
alright, and i get to but i dont know what kind of two they have weird answer choices for me?
2*
is it +-2 or the square root =-2?
\[8x^2 - 5 = 11\] Add 5 to both sides of the equation to isolate x: \[8x^2-5(+5) = 11 +5\] The 5 on the left will cancel out leaving you with:\[8x^2 = 16\] \[x^2 = 16\] \[x = \sqrt{16}\] \[x = 4\]
Oh wait, I skipped a step; sorry. x.x It should be....
alright but all my choices have to do with a 2
wait no they dont so its 4?
No, I skipped a step; so ignore that for now.
\[8x^2 = 16\] \[x^2 = 2\] \[x = \sqrt{2}\]
Is it asking for two answers, though?
nope just one thank you so much its the squear root of two?
\[\pm \sqrt{2}\] ; you can't have a negative under a radical; since that would be a completely different concept of Algebra. :p
what about this one: What constant term should be used to complete the square? x^2 - 5x + _____ = 7
Let me see if I can remember this, been a while. :)
\[x^2-5x+ ? = 7\] What you're going to take the middle term of your polynomial divide it by 2 and square it, if that makes sense.
So the first step:
\[x^2-5x+\frac{ 10 }{ 4 }=7\]
alright ...got it
Since you added 10/4 on the left, you're going to same on the right. \[x^2-5x+\frac{ 10 }{ 4 }=7+\frac{ 10 }{ 4 }\]
x^2-5x+10/4 = 38/4 \[(x-\frac{ 5 }{ 2 })^2=\frac{ 38 }{ 4 }\]
@exzaireah -- Is this helping at all? :x
yes im understandin just not any of my answer choices.
What are the choices given to you? Just to make sure I'm doing this correctly; I don't want to confuse you by doing the wrong thing.
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