Melissa_Something
  • Melissa_Something
Solve the inequality :(
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Melissa_Something
  • Melissa_Something
\[x^4-32x^2-144>0\]
Melissa_Something
  • Melissa_Something
I don't even know where to start..
Loser66
  • Loser66
let t = x^2 to take it down

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Melissa_Something
  • Melissa_Something
Is that substituting?
Loser66
  • Loser66
yup
Melissa_Something
  • Melissa_Something
What do we do with it? :s
Loser66
  • Loser66
do it first and I guide you more.
Melissa_Something
  • Melissa_Something
\[t^2-3xt-144>0\]
Loser66
  • Loser66
t^2 -32t -144 >0
Melissa_Something
  • Melissa_Something
Oh yeah typo
Loser66
  • Loser66
ok, now solve for quadratic to get t
Melissa_Something
  • Melissa_Something
um for x do I change it to t?
Loser66
  • Loser66
we will be back to x later. Now, just solve for t
Melissa_Something
  • Melissa_Something
for a i just put t? it looks weird lol
Loser66
  • Loser66
|dw:1444598561849:dw|
Melissa_Something
  • Melissa_Something
-32 = -1,024 .. am I doing this right?
Loser66
  • Loser66
complete the square for the left hand sides.
Melissa_Something
  • Melissa_Something
Oh so I have to solve this!! Okay so for the den of 2a I do 2^2? Or 2x^2
Loser66
  • Loser66
|dw:1444598640614:dw|
Loser66
  • Loser66
solve for t, then plug back to t = x^2 to solve for x. I let you finish the stuff.
Melissa_Something
  • Melissa_Something
Okay, but where did you get 16
Loser66
  • Loser66
From " complete the square" procedure. Rule: 32/2 =16, hence 16 is the number I need to + its square and then - its square.
Loser66
  • Loser66
|dw:1444599011779:dw|
Melissa_Something
  • Melissa_Something
So I didnt need to plug this into a quadratic?
Loser66
  • Loser66
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?
Melissa_Something
  • Melissa_Something
I just dont know what I was supposed to do first
Loser66
  • Loser66
Ok, I solve it for you as sample. You follow each step and see the reason why we have to do it.
Melissa_Something
  • Melissa_Something
Okay, Im just not good at math :(
Loser66
  • Loser66
\( t^2 -32 t- 144 > 0\\t^2 -32t +16^2 -16^2 -144 >0\\(t-16)^2 > 400 \\(t-16)^2 >20^2\)
Loser66
  • Loser66
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
Loser66
  • Loser66
If both are >0 , that is x-6 >0, x > 6 AND x +6 >0, x > -6 , combine them, you just have x > 6
Melissa_Something
  • Melissa_Something
Woah :o, but how did you get 400?
Loser66
  • Loser66
-16^2 -144 = -400 , move to other sides to get 400|dw:1444599855908:dw|
anonymous
  • anonymous
\[x^2>36,\left| x \right|>6,x<-6~ or~ x>6\]
Loser66
  • Loser66
oh, maybe you are right @surjithayer . I am so careful and solve it on the "order space". hehehe... Although it is flawless but a little bit ...overload to the Asker.
Melissa_Something
  • Melissa_Something
Yeah lol I havent even gotten to derivatives yet
Loser66
  • Loser66
No, it is just algebra.
Loser66
  • Loser66
But I was taught that way.
Loser66
  • Loser66
ok, you can take @surjithayer result. I think it is good enough.
Melissa_Something
  • Melissa_Something
I dont even know how @surjithayer got to that point :(
anonymous
  • anonymous
x^2-36>0 x^2>6^2 \[\left| x \right|>6\]
Melissa_Something
  • Melissa_Something
It was that easy
anonymous
  • anonymous
how Loser 66 was solving was correct also But a little longer
anonymous
  • anonymous
Take case 2 also and solve it
anonymous
  • anonymous
in this case you will get imaginary solution only.
Melissa_Something
  • Melissa_Something
The answer is (-\[(-\infty , -6) or (6, \infty) \] but how on earth do I get there? :(
anonymous
  • anonymous
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|
anonymous
  • anonymous
one is in Roaster form, and one is set Builder form.
Melissa_Something
  • Melissa_Something
Sorry Open Study didnt want to load ugh @surjithayer
Melissa_Something
  • Melissa_Something
Wait, um ugh im sorry but where did you get 36? @surjithayer
anonymous
  • anonymous
Loser66 solved x^2-36>0
anonymous
  • anonymous
\[\left| t-16 \right|>20 \] either t-16>20 t>20+16 or t>36 x^2>36 ......now further i have solved. or t-16 <-20 t<-20+16 t <-4 \[x^2< -4\] which is impossible because square of a real number is never negative. Hence rejected.

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