basic algebra and arithmetic "marathon". (1 Viewer)

[ohexploitable]

It's 2 root 3 and -root 3.

Bingo!

 

[ohexploitable]

are you sure.. i didnt get that

exploit wahts the answer


i got 11/2root3 and -6/2root3
using quadratic formula....

*facepalm*

Rationalise it lol.

 

[ohexploitable]

c'mon... someone else post a question...

 

[hscishard]

You said it

Will the world really end in 2012?

 

[Amogh]

I'll post a question tomorrow
If i remember

 

[cutemouse]

Find x: |x - 1| = -5

 

[thongetsu]

no solutions

 

[ohexploitable]

Find x: |x - 1| = -5

There are no solutions.

 

[ohexploitable]

x= -4 , 6

lol, absolute values can't be -ve so there are no solutions!

 

[thongetsu]

shit i didnt see the -ve sign

 

[ohexploitable]

New Question:

|t+2|+|3t-1|<5

 

[hscishard]

Find x: |x - 1| = -5

God. Thats like the third time you've posted that question.

 

[bouncing]

New Question:

|t+2|+|3t-1|<5

|t+2|<5-|3t-1|
t^2+4t+4<5-(9t^2-6t+1)
10t^2-2t<0
2t(5t-1)<0
therefore 0<t<1/5

 

[mirakon]

|t+2|<5-|3t-1|
t^2+4t+4<5-(9t^2-6t+1)
10t^2-2t<0
2t(5t-1)<0

therefore 0<T<1 p quote]< 5[>
doesn't look right to me....................

Don't you have to squarethe whole RHS instead of just what's in the absolute?

 

[hscishard]

New Question:

|t+2|+|3t-1|<5

t > -1
t < 1

Or, t is lesser than 1 but greater than -1

 

[ohexploitable]

t > -1
t < 1

Almost...

-1 < t < 1

 

[ohexploitable]

|t+2|<5-|3t-1|
t^2+4t+4<5-(9t^2-6t+1)
10t^2-2t<0
2t(5t-1)<0
therefore 0<t<1/5

Umm...

 

[hscishard]

Ummm you both used a very difficult way. You guys didn't sqaure it right.

5-|3t-1| sqaured is this!
(5-|3t-1|) sqaured.

Better use the case method

 

[hscishard]

Almost...

-1 < t < 1

Same thing Lol. How did you write it that way? It just fucking shows -1.

 

[hscishard]

-1 < t < 1

Awsome