Solve x^2-2x-3

Question

Solve  x^2-2x-3<0

1)  x<-1 
2)  x>3 
3)  -1<x<3

A) 1 only
B) 2 only
C) 1 & 2 only
D) 3 only
E) 1, 2, & 3

varunbawa · Answer

The answer here is D.
If you'll factorize the equation then (x-3)(x+1) < 0
This says that one of them is -ve and one is +ve, obviously x+1 > x-3 in all cases so we can say that
1. x+1>0 i.e. +ve => x>-1
2. x-3<0 i.e. -ve => x<3

Thus -1<x<3 is the solution.

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CharlieharpeR · Answer

Factorising the quadratic equation, we get :
x^2 - 2x - 3 => (x+1)(x-3) < 0

Now either (but not both) of these factors can be negative, only then the product will be negative.
Case 1: x+1 < 0 and x-3 > 0 => x<-1 and x>3 which cannot simultaneously be satisfied.
Case 2: x+1 > 0 and x-3 <0 => x>-1 and x<3 => Option D.

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TheUltimateWinner · Answer

hi  CharlieharpeR ,
Could you explain a little bit the  highlighted  part?
Thanks__

CharlieharpeR · Answer

Ok so like I said, for case1, x<-1 and x>3, both these conditions must be true at the same time. Now visualise a number line. All values of x satisfying x<-1 will lie to the left of -1 whereas all values of x satisfying x>3 will lie to the right of 3. Clearly these two intervals do not overlap. So there is no value of x that lies in both these areas and satisfies both conditions simultaneously.

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EgmatQuantExpert · Answer

Solution

Given:  
 • An inequality of the form  x^2 – 2x – 3 < 0

To find:  
 • The option choice, that explains the possible value of x

Approach and Working:  
We can start solving the inequality by factorizing the expression in the following way:
 •  x^2 – 2x – 3 < 0 
Or,  x^2 – 3x + x – 3 < 0 
Or,  x (x – 3) + 1 (x – 3) < 0 
Or,  (x – 3) (x + 1) < 0

Now, at x = -1 and x = 3, the equation becomes 0, and if we plot these two points on the number line, we get total 3 regions to consider:
 https://e-gmat.com/blogs/wp-content/uploads/2019/02/inequalities-1.png

We can check whether the inequality holds true in any specific region, by replacing x with any value belonging to that specific region.
 Region 1: 
 • If x = -2, (x – 3) = negative and (x + 1) = negative
Hence, (x – 3) (x + 1) = negative x negative = positive
But, original expression wise this should be negative.
Hence, Region 1 does not have feasible solutions of x.

Region 2: 
 • If x = -0, (x – 3) = negative and (x + 1) = positive
Hence, (x – 3) (x + 1) = negative x positive = negative
Also, original expression wise this should be negative.
Hence, Region 2 does have feasible solutions of x.

Region 3: 
 • If x = 4, (x – 3) = positive and (x + 1) = positive
Hence, (x – 3) (x + 1) = positive x positive = positive
But, original expression wise this should be negative.
Hence, Region 3 does not have feasible solutions of x.

https://e-gmat.com/blogs/wp-content/uploads/2019/02/inequalities-2.png

Therefore, we can say that feasible values of x exist in the range: -1 < x < 3
Hence, the correct answer is option D.

Answer: D

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-Algebra-e1533292163516.png

chetan2u · Answer

Surely case 3 AND both of case 1 and case 2 have nothing in common, so if 3 then none of 1 and 2 and if 1 or 2, then surely not 3. So, let us check a value in the range of 3.. 3) -10 would mean range is OUTSIDE the critical points, so x<-1 and x>3 These are two of many quick observations and tricks that can save a lot of time, so try to get a hang of such things.

Kinshook · Answer

x^2-2x-3<0 
  x^2-3x+x-3<0 
  x(x-3)+x-3<0 
  (x+1)(x-3)<0 
 -1<x<3

IMO D

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Solve x^2-2x-3<0

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