Answer: -1/2 < y < 1
In interval notation, that would look like (-1/2, 1).
Make sure to not confuse this with ordered pair notation which unfortunately looks identical to interval notation if you use curved parenthesis for both endpoints.
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Explanation:
2y^2 < y+1
2y^2 - y < 1
2y^2 - y - 1 < 0
2y^2 - 2y + y - 1 < 0
(2y^2 - 2y) + (y - 1) < 0
2y(y - 1) + 1(y - 1) < 0
(2y + 1)(y - 1) < 0
Solve (2y + 1)(y - 1) = 0 to get
(2y + 1)(y - 1) = 0
2y+1 = 0 or y-1 = 0
2y = -1 or y = 1
y = -1/2 or y = 1
These roots will help set up the boundaries of the intervals. Draw out a number line and plot the points -1/2 and 1 on the number line
Note how I have color-coded three regions A, B, C
Region A = interval to the left of -1/2
Region B = interval between -1/2 and 1
Region C = interval to the right of 1
Set up a sign chart as shown below.
The way I filled this chart out was plugging the representative y values for each region into each factor, then noting the sign (positive or negative). For instance, plug in y = -1 into each factor (2y+1) and (y-1) and you should get a negative result. We dont care what the actual value is. All we need is the sign of the results. When you multiply two negative values, you get a positive value. So thats why (2y + 1)(y - 1) is positive when -infinity < y < -1/2, or simply when y < -1/2. So region A is not part of the final answer. The other cells of the table are filled out in a similar way. I've highlighted the only region in which we get a negative value which is region B.
So this is why the answer is -1/2 < y < 1
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A similar problem is solving
2x^2 < x+1
which turns into
2x^2-x-1 < 0
Let f(x) = 2x^2-x-1
We want to find x values such that f(x) < 0
We can find them through graphing
The region in blue represents the solution set, which is from x = -1/2 to x = 1, excluding both endpoints. In this blue region, the red f(x) curve is below the x axis.