Another inequality problem

Question

Solve |(x)(x) + 3x| + (x)(x) -2 >= 0

venksune · Answer

two cases
case1: x^2 + 3x + x^2 - 2 >= 0
=>2x^2=3x-2>= 0
=>(2x-1)(x+2) >= 0
=>x>=1/2 or x>=-2

case2: -(x^2+3x) +(x^2+2) >=0
=>-3x>= 2
=>x>=-2/3

so, essentially the set is -2 to 1/2

Dharmin · Answer

Really good question

Some changes to venksune's post: two cases case1: x^2 + 3x + x^2 - 2 >= 0 =>2x^2+3x-2>= 0 =>(2x-1)(x+2) >= 0 So either x>=1/2 AND x>=-2 implied X SHOULD BE > 1/2 (both brackets are positive) or x<=1/2 AND x<=-2 implied X SHOULD BE < -2 (both brackets are negative so atlast whole equation is positive) case2: -(x^2+3x) + x^2 - 2 >=0 =>-3x>= 2 Multiply with (-1) to change (-3x) to (+3x), which will change the inequality =>3x<= -2 => x<= (-2/3) COMBINING BOTH THE CASES, SET SHOULD BE -INFINITE TO -2/3 and 1/2 to +INFINITE (apply zero and test) :idea:

it would be interesting to know OA and Explanation, thanks Dharmin

smcgrath12 · Answer

Right on. There are 2 cases here:

1) x^2 + 3x + x^2 - 2 >= 0

Equate the above equation to 0 and solve for x. We get x=1/2 and x=-2. Mark those points on a number line. These will be your boundary points. Now, as the above quadratic equation is greater (and equal) to zero, the solution for x will lie outside the boundary points. That give us, x>=1/2 and x<=-2.

2) -(x^2 + 3x) + x^2 -2 >= 0

Solving that, we get x<=-2/3.

Combining the above two, we get x <= -2/3 and x >= 1/2.

My only question is, when we combine the two, shouldn't we consider the more limiting condition of x <= -2 and not x <= -2/3 ???

Dharmin · Answer

Answer: More Limiting condition when AND is there between two condition. (A subset B) All posible cases when OR condition. ( A union B) (e.g. consider two sub cases of CASE - 1, we have AND between two condiontion and so we took more limiting) x>=1/2 AND x>=-2 implied X SHOULD BE >= 1/2 However, we included condition of both the cases CASE - 1 (x>1/2) and CASE - 2(x < -2/3) bcoz we have OR condition there and not AND. Case - 2 has OR condition so add its favorable results -2

PracticeMore · Answer

Here we have just two cases :

Case 1 : x^2 + 3x > 0 and Case 2: x^2 + 3x < 0

Case 1 : x^2 + 3x + x^2 - 2 >= 0

or 2*x^2 + 3x - 2 >=0 or (x+2)(2x-1) >=0

so, either x+2 >=0 , 2x-1 >=0
or x+2 <=0 , 2x-1 <= 0

so either x >=- 2 , x >=1/2
or x <= -2 , x <= 1/2

==> x >= 1/2 or x <= -2 ----- (I)

But, x^2 + 3x > 0 , so x (x+3) >0

so x > 0, x >-3 == > x >0 ---- (II)
or x < 0, x < - 3 ==> x < -3 ---- (III)

So from I , II and III , x > 1/2 or x < -3

Case 2 : : -x^2 - 3x + x^2 - 2 <= 0

or - 3x - 2 >=0 or 3x + 2 <= 0

so, x <= -2/3 ---- (I)

But, x^2 + 3x< 0 , so x (x+3) < 0

so x < 0, x >-3 ---- (II)
or x > 0, x < - 3 ---- (III)

So from I , II and III , -3 < x <= -2/3 or x < -3

so from case 1 :
x > 1/2 or x < -3

and from case 2 :
-3 < x <= -2/3 or x < -3

so the solution should be

x < -3 , -3 < x <=-2/3 , and x > 1/2

PracticeMore · Answer

sorry it will be 

x = 1/2 (I missed the equality sign from 1st case)

Another inequality problem

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