Solving the inequality using the test point method

jhawk555

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Sep 26, 2006
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I am not very good at these and was hoping if someone could check my work.

x(squared) - 10 is less than or equal to -3x

I changed above to x(squared) +3x -10
(x+5)(x-2)
x=-5
x=2

I chose test points -6, 0 and 4

1. (-1,-8) False
2. (5, -2) False
3. (-1, 2) True

My solution: x less than equal to -5 or x greater than or equal to 2.

Did I do this right?
 
jhawk555 said:
x(squared) - 10 is less than or equal to -3x

I changed above to x(squared) +3x -10
This should be:

. . . . .x^2 + 3x - 10 < 0

You don't want to lose the inequality sign.

jhawk555 said:
(x+5)(x-2)
x=-5
x=2
I will guess that this means the following:

. . . . .I found the zeroes (the endpoints of the intervals)
. . . . .by switching to the related equality, and solving:

. . . . . . .x^2 + 3x - 10 = 0
. . . . . . .(x + 5)(x - 2) = 0
. . . . . . .x + 5 = 0 or x - 2 = 0
. . . . . . .x = -5 or x = 2

. . . . .So my intervals are (-infinity, -5), (-5, 2), and (2, infinty).

If so, then your method is correct, as are the intervals.

jhawk555 said:
I chose test points -6, 0 and 4
Those are good points. Zero is almost always a nice number to work with, and picking smaller number for the other intervals can simplify your job, also.

jhawk555 said:
1. (-1,-8) False
2. (5, -2) False
3. (-1, 2) True
I'm sorry, but I don't know what you mean by the above. How do these points relate to the x-values you picked (namely, -6, 0, and 4) or with the evaluated values of the quadratic?

jhawk555 said:
My solution: x less than equal to -5 or x greater than or equal to 2.
If you look at the graph of y = x^2 + 3x - 10, you will see that the parabola is below the x-axis (that is to say, the value of the quadratic is less than zero) between the two zeroes of the quadratic. So it is the other interval which is the solution.

Eliz.
 
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