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I don't even know where to start..
let t = x^2 to take it down
Is that substituting?
What do we do with it? :s
do it first and I guide you more.
t^2 -32t -144 >0
Oh yeah typo
ok, now solve for quadratic to get t
um for x do I change it to t?
we will be back to x later. Now, just solve for t
for a i just put t? it looks weird lol
-32 = -1,024 .. am I doing this right?
complete the square for the left hand sides.
Oh so I have to solve this!! Okay so for the den of 2a I do 2^2? Or 2x^2
solve for t, then plug back to t = x^2 to solve for x. I let you finish the stuff.
Okay, but where did you get 16
From " complete the square" procedure. Rule: 32/2 =16, hence 16 is the number I need to + its square and then - its square.
So I didnt need to plug this into a quadratic?
You did. But the way to solve this problem is not just this, you have many many steps to get the answer. I know it is not easy but no choice for you. Take steps. Understand each step first. So far so good now?
I just dont know what I was supposed to do first
Ok, I solve it for you as sample. You follow each step and see the reason why we have to do it.
Okay, Im just not good at math :(
\( t^2 -32 t- 144 > 0\\t^2 -32t +16^2 -16^2 -144 >0\\(t-16)^2 > 400 \\(t-16)^2 >20^2\)
Hence \(t -16 > 20 ~~or ~~t-16 <-20 \) Case 1: if t -16 >20 , then t -16 -20 > 0 . That is t -36 >0 Recall \( t = x^2\) , hence \(x^2 -36 > 0\) You can't solve directly from here. You MUST derive like \(x^2 -36 > 0\\(x-6)(x + 6 ) >0 \) For this case, both (x-6) and (x+6) must be the same sign. That is if (x-6) >0, (x+6) >0 also. Because (+) *(+) >0 if (x-6) <0, then (x+6) <0 also, since (-)*(-) = positive >0
If both are >0 , that is x-6 >0, x > 6 AND x +6 >0, x > -6 , combine them, you just have x > 6
Woah :o, but how did you get 400?
-16^2 -144 = -400 , move to other sides to get 400|dw:1444599855908:dw|
\[x^2>36,\left| x \right|>6,x<-6~ or~ x>6\]
Yeah lol I havent even gotten to derivatives yet
No, it is just algebra.
But I was taught that way.
x^2-36>0 x^2>6^2 \[\left| x \right|>6\]
It was that easy
how Loser 66 was solving was correct also But a little longer
Take case 2 also and solve it
in this case you will get imaginary solution only.
The answer is (-\[(-\infty , -6) or (6, \infty) \] but how on earth do I get there? :(
x<-6 or x>6 means \[\left( -\infty,-6 \right) U \left( 6,\infty \right)\] these both are same only way of writing is different. |dw:1444600690282:dw|
one is in Roaster form, and one is set Builder form.