Quiz20MorningSolution

Quiz20MorningSolution - Morning Quiz #20 Solution Mean: 5,...

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Unformatted text preview: Morning Quiz #20 Solution Mean: 5, High: 10, Low: 1 1. Solve the inequalities and express the solution set in interval notation. a. ( ) Solving nonlinear inequalities: 1. Set one side of the inequality to zero. Subtract 4 from both sides of the inequality so that the right-hand side is zero. ( ) 2. Combine like terms, then factor the other side. Distribute the factor of to ( ). Factor the trinomial. ( ) ( ( )( ) ) 3. Find the values that make each factor zero (including restrictions). 4. Set-up a number line using the values from step 3. 5. Find the intervals that satisfy the inequality by using test values. ( ∞, )( , ,∞ 2 2 ( )( ) ) ( )( ) ( )( ) )( ) Our inequality is ( , so the expression on the left needs to be negative or equal to zero. We can see from the chart that the expression will be negative on the interval from to ; since we also want the expression to equal zero, we will include both endpoints in the interval. [ ,] b. 2 Solving nonlinear inequalities: 1. Set one side of the inequality to zero. The right-hand side is already zero. 2. Combine like terms, then factor the other side. There are no like terms to combine, and the numerator is already factored, so we only need to factor the denominator. 2 ( ) ( 2( )( ) 2) 2 ( )( 2) 3. Find the values that make each factor zero (including restrictions). The numerator is zero when 2. The denominator is zero when or when 2; keep in mind that these are restrictions, so and 2 cannot be included in the solution. 4. Set-up a number line using the values from step 3. 2 2 5. Find the intervals that satisfy the inequality by using test values. 2 ( ∞, 2 2 )( 2 2, 2 2) 2, ,∞ 6 ( )( ) Our inequality is ( ( )( ) )( ( )( ) ) ( )( ) , so the expression on the left needs to be equal to zero or positive. We can see from the chart that the expression will be positive on the interval from 2 to 2 and on the interval from to ∞. We determined in step 3 that the expression is zero when 2, so we will include that value in our solution. Remember that we cannot include 2 or in our solution because those values make the denominator zero; that is why we use parentheses instead of brackets on those endpoints. ( ,] ( , ∞) ...
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