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23 Jul 2020, 21:49
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What is the range of X ?
when \( x( x^2-4 ) >0 \)
when \( x( x^2-4 ) >0 \)
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Hi,
The answer would be -2<x<0 and x>2
How to solve is as below
from x( x^2-4 ) >0 gives x<0 or x<2 and x>-2
Verifying which one of x>0 and x>2 fits
Lets take x=0, the inequality becomes 0>0 WRONG
if x=-1, the inequality becomes 3>0 CORRECT.
if x=3, the inequality becomes 15>0 CORRECT
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Updated on: 17 Jul 2021, 09:24
Originally posted by BrushMyQuant on 23 Jul 2020, 22:22.
Last edited by BrushMyQuant on 17 Jul 2021, 09:24, edited 2 times in total.
Last edited by BrushMyQuant on 17 Jul 2021, 09:24, edited 2 times in total.
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There are three ways of solving inequality problems
Check out this post to learn How to Solve Inequality Problems
Method 1: Sine Wave / Wave / Wavy Method
We will find out the points on the number line assuming the inequality was = 0
So, \( x( x^2-4 ) =0 \)
=> x = 0, -2, +2
So, we will plot the three points on the number line and plot the sign curve as follows:
- Start from the right top and go down at first solution (2)
- Come up in second solution (0)
- Go down in third solution (-2)
Now we are going to write (+) and (-) as follows:
- Any point above the number line and below the sine curve we will write (+)
- Any point below the number line and above the sign curve we will write (-)
Since the question is >0 so we are going to pick all the range in which we have (+)
So, answer is x > 2 and -2 < x < 0
Method 2: Substitution
We will plot the three points as we did above
Now, in each and every range (as shown in 4 colors above), we will go ahead and substitute values in all these ranges as follows
1. x > 2, Let's try x=3 and see if \( x( x^2-4 ) >0 \) is satisfied
\( 3( 3^2-4 ) >0 \) => 3 * 5 = 15 > 0
So, x > 2 is one solution range
2. 0 < x < 2, Let's try x=1 and see if \( x( x^2-4 ) >0 \) is satisfied
\( 1( 1^2-4 ) >0 \) => 1 * -3 = -3 < 0
So, 0 < x < 2 is NOT a solution range
3. -2 < x < 0, Let's try x=-1 and see if \( x( x^2-4 ) >0 \) is satisfied
\( -1(- 1^2-4 ) >0 \) => -1 * -3 = 3 > 0
So, -2< x < 0 is one solution range
4. x < -2, Let's try x = -3 and see if \( x( x^2-4 ) >0 \) is satisfied
\( -3( -3^2-4 ) >0 \) => -3 * 5 = -15 < 0
So, x <- 2 is NOT a solution range
So, answer is x > 2 and -2 < x < 0
Method 3: Algebra
We can consider \( x( x^2-4 ) >0 \) as
AB > 0, and we know that there will be two cases
Either Both A and B > 0
Or Both A and B < 0
Case 1: Both A and B are > 0
x > 0 and \(x^2\) - 4 > 0
\(x^2\) - 4 > 0
=> x > 2 or x < -2
We know that x > 0
=> Solution is x > 2
Case 2: Both A and B are < 0
x < 0 and \(x^2\) - 4 < 0
\(x^2\) - 4 < 0
=> -2 < x < 2
We know that x < 0
=> Solution is -2 < x < 0
So, Solution is -2 < x < 0 or x > 2
Hope it helps!
Watch the following video to learn how to solve inequality problems
Check out this post to learn How to Solve Inequality Problems
Method 1: Sine Wave / Wave / Wavy Method
We will find out the points on the number line assuming the inequality was = 0
So, \( x( x^2-4 ) =0 \)
=> x = 0, -2, +2
So, we will plot the three points on the number line and plot the sign curve as follows:
- Start from the right top and go down at first solution (2)
- Come up in second solution (0)
- Go down in third solution (-2)
Now we are going to write (+) and (-) as follows:
- Any point above the number line and below the sine curve we will write (+)
- Any point below the number line and above the sign curve we will write (-)
Since the question is >0 so we are going to pick all the range in which we have (+)
So, answer is x > 2 and -2 < x < 0
Method 2: Substitution
We will plot the three points as we did above
Now, in each and every range (as shown in 4 colors above), we will go ahead and substitute values in all these ranges as follows
1. x > 2, Let's try x=3 and see if \( x( x^2-4 ) >0 \) is satisfied
\( 3( 3^2-4 ) >0 \) => 3 * 5 = 15 > 0
So, x > 2 is one solution range
2. 0 < x < 2, Let's try x=1 and see if \( x( x^2-4 ) >0 \) is satisfied
\( 1( 1^2-4 ) >0 \) => 1 * -3 = -3 < 0
So, 0 < x < 2 is NOT a solution range
3. -2 < x < 0, Let's try x=-1 and see if \( x( x^2-4 ) >0 \) is satisfied
\( -1(- 1^2-4 ) >0 \) => -1 * -3 = 3 > 0
So, -2< x < 0 is one solution range
4. x < -2, Let's try x = -3 and see if \( x( x^2-4 ) >0 \) is satisfied
\( -3( -3^2-4 ) >0 \) => -3 * 5 = -15 < 0
So, x <- 2 is NOT a solution range
So, answer is x > 2 and -2 < x < 0
Method 3: Algebra
We can consider \( x( x^2-4 ) >0 \) as
AB > 0, and we know that there will be two cases
Either Both A and B > 0
Or Both A and B < 0
Case 1: Both A and B are > 0
x > 0 and \(x^2\) - 4 > 0
\(x^2\) - 4 > 0
=> x > 2 or x < -2
We know that x > 0
=> Solution is x > 2
Case 2: Both A and B are < 0
x < 0 and \(x^2\) - 4 < 0
\(x^2\) - 4 < 0
=> -2 < x < 2
We know that x < 0
=> Solution is -2 < x < 0
So, Solution is -2 < x < 0 or x > 2
Hope it helps!
Watch the following video to learn how to solve inequality problems
