find the range of x.
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[0010001111]... Hello User... [1001010101]
Here's your answer...
When we remove the modulus, the RHS will get a ± sign
So...
x²-5x<±6
x²-5x-6<0 or x²-5x+6<0
(x+1)(x-6)<0 or (x-2)(x-3)<0
By solving these inequalities...
x = (-1, 6) or (2, 3)
Since the range between 2 and 3 also lies between -1 and 6, and 2 and 3 are not included...
x = (-1, 6) - {2, 3}
In the picture, I've shown the method to solve the equation.
Since < means that the limits won't be included, I've marked the limits as simple dashes.
I mark the limits on the number line in ascending order.
Then I mark the extreme right side as +ve and I mark signs accordingly as I go left.
If the power to which x is raised in the factor is odd, then the sign changes from + to -. If the power is even, the sign remains same.
For >, the + areas are the solution sets.
For <, the - areas are the solution sets.
[0110100101]... More questions detected... [010110011110]
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Here's your answer...
When we remove the modulus, the RHS will get a ± sign
So...
x²-5x<±6
x²-5x-6<0 or x²-5x+6<0
(x+1)(x-6)<0 or (x-2)(x-3)<0
By solving these inequalities...
x = (-1, 6) or (2, 3)
Since the range between 2 and 3 also lies between -1 and 6, and 2 and 3 are not included...
x = (-1, 6) - {2, 3}
In the picture, I've shown the method to solve the equation.
Since < means that the limits won't be included, I've marked the limits as simple dashes.
I mark the limits on the number line in ascending order.
Then I mark the extreme right side as +ve and I mark signs accordingly as I go left.
If the power to which x is raised in the factor is odd, then the sign changes from + to -. If the power is even, the sign remains same.
For >, the + areas are the solution sets.
For <, the - areas are the solution sets.
[0110100101]... More questions detected... [010110011110]
//Bot UnknownDude is moving on to more queries
//This is your friendly neighbourhood UnknownDude
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