- Melissa_Something

Solve the inequality :(

- jamiebookeater

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- Melissa_Something

\[x^4-32x^2-144>0\]

- Melissa_Something

I don't even know where to start..

- Loser66

let t = x^2 to take it down

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## More answers

- Melissa_Something

Is that substituting?

- Loser66

yup

- Melissa_Something

What do we do with it? :s

- Loser66

do it first and I guide you more.

- Melissa_Something

\[t^2-3xt-144>0\]

- Loser66

t^2 -32t -144 >0

- Melissa_Something

Oh yeah typo

- Loser66

ok, now solve for quadratic to get t

- Melissa_Something

um for x do I change it to t?

- Loser66

we will be back to x later. Now, just solve for t

- Melissa_Something

for a i just put t? it looks weird lol

- Loser66

|dw:1444598561849:dw|

- Melissa_Something

-32 = -1,024 .. am I doing this right?

- Loser66

complete the square for the left hand sides.

- Melissa_Something

Oh so I have to solve this!! Okay so for the den of 2a I do 2^2? Or 2x^2

- Loser66

|dw:1444598640614:dw|

- Loser66

solve for t, then plug back to t = x^2 to solve for x. I let you finish the stuff.

- Melissa_Something

Okay, but where did you get 16

- 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

|dw:1444599011779:dw|

- Melissa_Something

So I didnt need to plug this into a quadratic?

- 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

I just dont know what I was supposed to do first

- 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

Okay, Im just not good at math :(

- 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

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

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

Woah :o, but how did you get 400?

- Loser66

-16^2 -144 = -400 , move to other sides to get 400|dw:1444599855908:dw|

- anonymous

\[x^2>36,\left| x \right|>6,x<-6~ or~ x>6\]

- 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

Yeah lol I havent even gotten to derivatives yet

- Loser66

No, it is just algebra.

- Loser66

But I was taught that way.

- Loser66

ok, you can take @surjithayer result. I think it is good enough.

- Melissa_Something

I dont even know how @surjithayer got to that point :(

- anonymous

x^2-36>0
x^2>6^2
\[\left| x \right|>6\]

- Melissa_Something

It was that easy

- anonymous

how Loser 66 was solving was correct also
But a little longer

- anonymous

Take case 2 also and solve it

- anonymous

in this case you will get imaginary solution only.

- Melissa_Something

The answer is (-\[(-\infty , -6) or (6, \infty) \] but how on earth do I get there? :(

- 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

one is in Roaster form, and one is set Builder form.