Division by 0 is not allowed, thus 1/0 is undefined. So the answer to your question is - no.
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Hi Gurpreet ,

Can you please explain how did you did you say from the second statement that x <- 3and x> -1 ?

The factors of the equation would be -1 and -3 how how do we know the sign for x ... (I have read many forms but can't understand)

Have you checked this post: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476 and this: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661-20.html#p1365336 ?
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Hi Bunuel,
Can I safely deduce the following

When (x+3)(x+1) = 0 , There are exactly two solutions

When (x+3) (x+1)> 0 , there are multiple solutions. Because by putting any positive value of x I can get >0

When (x+3) (x+1)<0, Either (x+3) or (x+1) has to be negative and obviously none can be Zero. In that case to make negative I can take only one value which is -2.

I don't want to concentrate on graphical method. Will the approach work for these types of questions always?
Please help

That's not correct.

(x + 3)(x + 1) > 0 means that x < -3 or x > -1.
(x + 3)(x + 1) < 0 means that -3 < x < -1.
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Hello Bunuel, I am not sure if I followed your explanation all that clearly. I was able to get to the factors of the equation but after that, you mention about "<" symbol and as a result you arrive at \(1<x<3\). How come? I tried to grasp your graphical approach but that did sit well either. Could you please elaborate? Thank you in advance!

Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html
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After spending long time on this topic, I have come to an easy solution for me. You may like it

(1) (x-3)(x-1)<0. Now use the logic. To get negative result Either (x-3) or (x-1) has to be negative and the other has to +ve

Here, (x-3) can't be positive, Because if x-3>0 then x>3 and if x>3 then x-1 will become +ve. So (x-3)(x-1) will become +ve.

Therefore, (x-3) has to be negative,x-3<0 or x<3 and and as I said earlier x-1>0 or x>1

Finally we get, 1<x<3. Sufficient because 2 is the only integer between 1 and 3.

(2) we don't need to find equation. Just by putting multiple positive values we can satisfy statement 2. So Not sufficient

Ans. A
Hopefully my thinking is correct. I expect expert reply. Can I get more such question to practice?

Given: \(x=integer\).

(1) \(x^2-4x+3<0\) --> \(1<x<3\)--> as \(x\) is an integer then \(x=2\). Sufficient.

(2) \(x^2+4x+3>0\) --> \(x<-3\) or \(x>-1\) --> multiple values are possible for integer \(x\). Not sufficient.

Answer: A.[/quote]

With reference to your statement \(1<x<3\) above, I calculated the range as \(x<1\) or \(x<3\)
What I did wrong?[/quote]

How to solve quadratic inequalities - Graphic approach.

\(x^2-4x+3<0\) is the graph of parabola and it look likes this:
Intersection points are the roots of the equation \(x^2-4x+3=0\), which are \(x_1=1\) and \(x_2=3\). "<" sign means in which range of \(x\) the graph is below x-axis. Answer is \(1<x<3\) (between the roots).

If the sign were ">": \(x^2-4x+3>0\). First find the roots (\(x_1=1\) and \(x_2=3\)). ">" sign means in which range of \(x\) the graph is above x-axis. Answer is \(x<1\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root).


This approach works for any quadratic inequality. For example: \(-x^2-x+12>0\), first rewrite this as \(x^2+x-12<0\) (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern).

\(x^2+x-12<0\). Roots are \(x_1=-4\) and \(x_1=3\) --> below ("<") the x-axis is the range for \(-4<x<3\) (between the roots).

Again if it were \(x^2+x-12>0\), then the answer would be \(x<-4\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root).

Hope it helps.[/quote]

I am fond of your explanations !!
++++kudos

Product of 2 terms <0 , then the variable lies between the given points.

(x-a) (x-b)<0, (considering a<b, in the number line)

1st Condition

Either x-a<0 i.e x<a or x-b>0 i.e x>b - NO common region

2nd Condition

Either x-a>0 i.e x>a or x-b<0 i.e x<b - a<x<b

Hope this helps.
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Hi Bunuel, how did you derive the range without using the parabola approach? I got the answer by picking numbers and though I got the correct answer, it was a bit time-consuming.

If x is an integer, what is the value of x?

(1) \(x^2 - 4x + 3 < 0\)

\((x-3) (x-1) < 0\)
\((x-3)\) and \((x-1)\) must have opposite signs.
We can conclude \(1 < x < 3\).
Since x is an integer, x must be 2. Sufficient.

(2) \(x^2 + 4x +3 > 0\)
(x+3)(x+1) > 0
(x+3) and (x+1) are positive.
Multiple answers are possible. x can be 4, 5, 6, etc. Insufficient.

Answer is A.

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(1) (x-1)(x-3)<0; Now putting the value on the Wavy curve

---------------\(+\)[/color--------------1--------[color=#00a651]\(-\)------3---\(+\)----------------

\(1<x<3\); as \(x\) is an integer. Sufficient.

(2) \((x+1)(x+3)>0\); Now putting the value on the Wavy curve


-1--\(-\)-- -1+

\(x>-1, \)or

\(x<-3;\) Insufficient.

The answer is A
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Really feeling better about inequalities / mods in the last couple of days. Everytime I see them I usually freak out.

x =?

(1) x^2 - 4x + 3 < 0

1. (x -1) (x-3) < 0 <----implies that (x-1) OR (x-3) is negative (but not both)
2. Find the critical points:
(x-1) <0 ---> x < 1
(x-3) <0 ---> x < 3
3. Find the region where the inequality is negative: 1 < x < 3 <---Notice that since there are no equal to sign that x MUST equal 2.

Sufficient

(2) x^2 + 4x +3 > 0
1. (x +1) (x+3) > 0 <----implies that (x+1) OR (x+3) are both negative or are both positive
2. Find the critical points:
(x+1) >0 ---> x > -1
(x+3) >0 ---> x > -3
3. Find the region where the inequality is positive: x < -3 or x > -1

Since x < -3 and x > -1 , there are multiple possible integer values for x

Insufficient.

Anytime you see quadratic equations on the exam, you probably have to factor. It's no different here.
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Given x is an integer.
(A) x^2-4x+3<0 can be re-written as (x-2)^2-1<0......Since the square term can't be less than zero, the value of x = 2. Sufficient.
(B) x^2+4x+3>0 can be re-written as (x+2)^2+1>0....Which is true for all values of x. Not Sufficient.

Hence, A is the Answer.