# 11.4 - 11.4 Nonlinear Inequalities in One Variable...

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Unformatted text preview: 11.4 Nonlinear Inequalities in One Variable Quadratic lnegualities: A quadratic inequality is an inequality that can be written so that one side is a quadratic expression and the other side is 0. Examples 3x2—4>0 -2x2—5x+7SO x2+4x-—620 2x2—3<O A solution of a quadratic inequality is a value of the variable that makes the inequality a true statement. Solving a Polynomial Inequality _1) Write the inequality in standard form, then solve the related equation. ' 2) Separate the number line into regions with the solutions from Step 1. 3) For each region, choose a test point and determine whether its value satisfies the original inequality. 4) The solution set includes all the regions whose test point value is a solution of the inequality. - If the inequality symbol is 2 or s, the values from Step 1 are solutions; ° If the inequalitysymbol is < or >, the values from Step 1 ' are not included. If we attempt to solve a quadratic inequality, such as 3x2 + 5x + 2 < 0, we are looking for values ofx that will make this a true statement. If we graph the quadratic equation y = 3x2 + 5x + 2, the points of the parabola that lie below the x—axis would provide values ofx where the y— value < 0. Hence, those values ofx would satisfy 3x2 + 5x + 2 < 0. Similarly, we could also use the graph to find the values ofx that satisfy the inequality 3x2 + 5X + 2 > O (the x-values of all points above the x—axis). The points on the graph above and below the x—axis are separated by points actually on the x-axis. These points would have values of x such that 3x2 + 5x+ 2 = 0. However, graphing a quadratic equation could be time consuming if you don’t have a computer or graphing calculator. Fortunately, it is not necessary to graph a quadratic inequality to solve this type of problem. We can construct a number line representing the x-axis and find the region(s) on the number line where the inequality is true. Solve the quadratic inequality 3x2 + 5x + 2 < 0. Bx‘ﬁsx +1 4 O 2 6 ‘ 3X \$1M +246 +-2_ 4 c) 3x (X+f)+2—(X+l) 4. o (3 X+le<+f ) 4.0 \——i) (—6-+9_)4o Edi) (—H) 40 ML} <0 i"al%€ _ I “Tu-L Pia X151 (3(1)+2_] [2+04o £6;ij [33443 ‘8' LB) 4:) lit 40 False. Tég-wL 1L X245 (”%“4)C“%+é)?o @”4)LO"H;)>O (54%) [57%) >0 (“{2} (#2)?” C—Wé)>o UN”) >0 “24 >0 H >0 35E tam TRUE;- Solve the polynomial inequality x(x — 6)(x + 2) 2 O. *3 (—3—4) (esp—DZ 0 —3, (4299026 “'CJUC‘D7/‘5 #25125 #7/0 RISP— (New; [Cw—r7") (3) 370 —— £57,- 0 rFafse ...
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11.4 - 11.4 Nonlinear Inequalities in One Variable...

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