Is x 1? (1) x^2 + x + 2 8 (2) 8(x – 4) 4(x – 2)

Question

Is  x > 1 ?

(1)  x^2 + x + 2 > 8

(2)  8(x – 4) > 4(x – 2)

DavidTutorexamPAL · Answer

Instead of solving explicitly, we'll look for simple numbers that contradict the given statements.
This is an Alternative approach.

(1)
Let's try a number that gives us a YES. If we make x very large, say x = 100, then 100^2+100+2 is definitely larger than 8.
Not let's try to contradict this by getting a NO. If we make x very small, say x = -100 then (-100)^2 - 100 + 2 is 10,000 -100 + 2 which is also larger than 8.
We have found both x > 1 and x < 1 which gives a correct statement (1) so this is not enough.
Insufficient!

(2)
Since it's hard to guess what numbers to pick, we'll first simplify a bit.
8x - 32 > 4x - 8
4x > 24
x > 6.
In this case, there is no need to pick numbers at all!
Sufficient.

(B) is our answer.

Tulkin987 · Answer

The first statement tells us that 
 x^2 + x + 2 > 8  or
 x^2 + x - 6 > 0 ,
 (x+3)(x-2)>0 .
So, from this, we know that   x>2  and  x<-3  . Insufficient.

If we solve the second inequality, then we end up with   x>6   which is sufficient.

Answer: B

mamun2871 · Answer

1) From 1 we get X>2 and X<-3. 
2) From 2 we get X>6.

Hence B is my answer.

axmanX · Answer

Hi,

Please explain

How did you get from
x+3>0 to x<-3

I couldn't understand it.

PKN · Answer

Question stem:- Is x>1 ?

St1:-  x^2 + x + 2 > 8 
Or,  x^2+x-6>0 
Or,  x^2-2x+3x-6>0 
Or,  x(x-2)+3(x-2)>0 
Or,  \left(x-2\right)\left(x+3\right)>0 
Cut off points:- x=2, -3
Applying wavy-curve method(figure enclosed), the region of the curve above horizontal axis(since the product is positive),
x<-3 , x>2
So, x may or mayn't be greater than 1.
Insufficient.

St2:-  8(x – 4) > 4(x – 2) 
Or, 8x-32>4x-8
Or, 8x-4x>-8+32
Or, 4x > 24
Or,  x> \frac{24}{4} 
Or, x > 6
Sufficient.

Ans. (B)

agv2501 · Answer

can someone please explain how we are moving the variable x across the inequalities without knowing the sign? How is it being assumed that X is positive and hence will not have an effect on the inequality while solving for it..? What am i missing..?

ueh55406 · Answer

Hey man, it's a common misconception. Yes you are right, we cannot move variables around w/o knowing the sign. so, always stick to that concept. what's happening here, or for any other question where you see this happening, is that variables are not being moved. we have only subtracted that variable from both side of the inequality.

here :

-3p-7 > p + 9  now we will not " move P to the right side" what we will do is subtract P both side

Subtracting +p both side (we get the following) (
 -3p-7 > p + 9

we're left with:

-4p-7 > 9  
Add 7 to both side
 -4p >16   
divide by -4 on both side:
 p < -4

Subtracting/adding the same value/variable (with the same sign) on both side of the inquality does not change the sign of the inequality

Is x > 1? (1) x^2 + x + 2 > 8 (2) 8(x – 4) > 4(x – 2)

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