#solving inequalities
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atomic bluff
Did you set it equal to 0 first of all?
Because it makes it a lot easier
Then (x-5)(x-3) = 0
No
Because 0/ something = 0
Then x = 5 or 3
So now we look at cases
We know that we can only have negative / negative, or positive / positive
Greater or equal to 5 would result in 0
Or some positive number / positive number
-1?
Well we look for the next zero
Which is x = 3
So now we look for values greater than 3 and less than three
Plugging in 4
We have to figure out the zeroes to find number around that
-4 and 1 aren’t zeroes
That makes the thing undefined
Now we have ranges
pulsar kiln
We can't solve the inequality x² - 8x + 15 >= 0 any further, but we can find the critical numbers and a sign chart:
Critical numbers:
x = 1 (from x - 5 = 0)
x = 4 (from x - 3 = 0)
x = -4 (from x + 4 = 0)
Sign chart:
(--)(+)(+)(+)(++)
The solution to the inequality is (-∞, -4] U [1, ∞).
atomic bluff
Plug in -1/2 for yourself
-1/2 > -1
Plug it in and see if it works
You don’t need the values just if it’s negative or positive
atomic bluff
We know that we can only have negative / negative, or positive / positive
Let’s assume the numerator is positive
For one case
We have x > 5
So (x-5) and (x-3) are both positive
The other case is that x < 3
So that both are negative
pulsar kiln
Square brackets are inclusive and parentheses are exclusive. The brackets go on an endpoint that is included and the parentheses go on an endpoint that is not included in the interval.
paper cosmosBOT
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