document.write( "Question 1150847: Solve Inequality: 2y^2 < y+1 \n" ); document.write( "
Algebra.Com's Answer #772376 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Answer: -1/2 < y < 1\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "In interval notation, that would look like (-1/2, 1). \n" ); document.write( "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.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "========================================= \n" ); document.write( "Explanation:\r \n" ); document.write( "\n" ); document.write( "2y^2 < y+1 \n" ); document.write( "2y^2 - y < 1 \n" ); document.write( "2y^2 - y - 1 < 0 \n" ); document.write( "2y^2 - 2y + y - 1 < 0 \n" ); document.write( "(2y^2 - 2y) + (y - 1) < 0 \n" ); document.write( "2y(y - 1) + 1(y - 1) < 0 \n" ); document.write( "(2y + 1)(y - 1) < 0\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Solve (2y + 1)(y - 1) = 0 to get \n" ); document.write( "(2y + 1)(y - 1) = 0 \n" ); document.write( "2y+1 = 0 or y-1 = 0 \n" ); document.write( "2y = -1 or y = 1 \n" ); document.write( "y = -1/2 or y = 1\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "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 \n" ); document.write( " ![]() \n" ); document.write( "Note how I have color-coded three regions A, B, C \n" ); document.write( "Region A = interval to the left of -1/2 \n" ); document.write( "Region B = interval between -1/2 and 1 \n" ); document.write( "Region C = interval to the right of 1\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Set up a sign chart as shown below. \n" ); document.write( " ![]() \n" ); document.write( "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.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So this is why the answer is -1/2 < y < 1\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A similar problem is solving \n" ); document.write( "2x^2 < x+1 \n" ); document.write( "which turns into \n" ); document.write( "2x^2-x-1 < 0 \n" ); document.write( "Let f(x) = 2x^2-x-1 \n" ); document.write( "We want to find x values such that f(x) < 0 \n" ); document.write( "We can find them through graphing \n" ); document.write( " ![]() \n" ); document.write( "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. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |