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# Solve x^2-2x-3<0

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VP
Joined: 23 Feb 2015
Posts: 1258

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20 Feb 2019, 13:28
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Difficulty:

45% (medium)

Question Stats:

64% (01:17) correct 36% (01:13) wrong based on 22 sessions

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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

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Intern
Joined: 22 May 2017
Posts: 17

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20 Feb 2019, 13:41
1
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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Joined: 19 Feb 2019
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20 Feb 2019, 14:18
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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VP
Joined: 23 Feb 2015
Posts: 1258

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20 Feb 2019, 14:28
CharlieharpeR wrote:
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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hi CharlieharpeR,
Could you explain a little bit the highlighted part?
Thanks__
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Intern
Joined: 19 Feb 2019
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20 Feb 2019, 15:13
1
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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20 Feb 2019, 20:13

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:

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.

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

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Re: Solve x^2-2x-3<0   [#permalink] 20 Feb 2019, 20:13
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