The solution set of o the quadratic equation x2 6x 5 0

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Unformatted text preview: y-coordinate is 0. y y>0 4 • The solution of the y = 0 0 associated "greater than" inequality is y<0 found at the points −4 on the graph that are also above the x-axis where the x2 − 6 x + 5 > 0 y-coordinate is x<1 greater than 0. y = x2 − 6 x + 5 > 0 y = x2 − 6x + 5 y = x2 − 6 x + 5 = 0 1 x 5 y = x2 − 6 x + 5 < 0 x2 − 6 x + 5 < 0 1<x<5 x2 − 6 x + 5 > 0 x>5 • The solution of the associated "less than" inequality is found at the points on the graph that are also below the x-axis where the y-coordinate is less than 0. The solution set of: o the quadratic equation x2 − 6x + 5 = 0 is { 1 , 5 } o the quadratic inequality x2 − 6x + 5 > 0 is { x | x < 1 or x > 5, x ∈ R } o the quadratic inequality x2 − 6x + 5 ≥ 0 is { x | x ≤ 1 or x ≥ 5, x ∈ R } o the quadratic inequality x2 − 6x + 5 < 0 is { x | 1 < x < 5, x ∈ R } o the quadratic inequality x2 − 6x + 5 ≤ 0 is { x | 1 ≤ x ≤ 5, x ∈ R } Unit 3: Day 7 notes - Quadratic Inequalities in One Variable Page 2 of 3 SOLVING QUADRATIC INEQUALITIES IN ONE VARIABLE ALGEBRAICALLY example: Solve the quadratic inequality x2 − 6x + 5 < 0 • We have seen from the graphical method that the key numbers for the solution are the roots of the associated equation; roots that can be found algebraically. Solve x2 − 6x + 5 = 0 algebraically; the roots are 1 and 5. • Place the roots on a number line; use closed circles if these numbers are included in the solution and open circles if these numbers do not satisfy the inequality. x 1 5 • These numbers break up the number line into regions. Test a value from within each region; if it satisfies the inequality, then all the numbers from that region will satisfy is as well. x 1 5 2 Test 0: (0) − 6(0) + 5 = 5 ≮ 0; no, not this region Test 2: (2)2 − 6(2) + 5 = −5 < 0; yes, mark the points in this region Test 6: (6)2 − 6(6) + 5 = 5 ≮ 0; no, not this region OR If the inequality is in factored form, (x − 1)(x − 5) < 0 : use Sign Analysis: Test numbers from each region, but just determine if the factors are positive of negative to determine sign of the product. Test 0: (x − 1)(x − 5) = (negative)(negative) = positive ≮ 0; no, not here Test 2: (x − 1)(x − 5) = (positive)(negative) = negative < 0; yes, here Test 6: (x − 1)(x − 5) = (positive)(positive) = positive ≮ 0; no, not here use Case Analysis: The product is negative so one of two cases must be true; either (case 1) x − 1 > 0 and x − 5 < 0 or (case 2) x − 1 < 0 and x − 5 > 0 . x > 1 and x < 5 or x < 1 and x > 5 1<x<5 or no such numbers • Write the solution set by describing the x-values from all the regions that satisfy the quadratic inequality. solution set = { x | 1 < x < 5, x ∈ R} Unit 3: Day 7 notes - Quadratic Inequalities in One Variable Page 3 of 3 exercise: Solve graphically first and then solve algebraically. a) x2 ≤ 9 b) x2 + 4x > 0 [Answer: {x| −3 ≤ x ≤ 3, x ∈R}] [Answer: {x| x < −4 or x > 0, x ∈R}] exercise: Solve algebraically. a) x2 − 3x − 54 ≥ 0 b) x2 + 2x − 1 < 0 [Answer: {x| x ≤ −6 or x ≥ 9, x ∈R}] [Answer: {x| −1 − 2 < x < 1 − 2 , x ∈R}] exercise: Solve algebraically. a) x2 > 0 b) x2 + 4x + 5 < 0 [Answer: {x| x ≠ 0, x ∈R}] [Answer: φ]...
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