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Warm Up Simplify the fraction: a) ππ π b) π+ππ π c) π+ππ π
I will determine the number and types of roots of a polynomial Warm Up Simplify the fraction: a) ππ π b) π+ππ π c) π+ππ π Solve using the quadratic formula π π π +ππβπ=π π₯= βπΒ± π 2 β4ππ 2π
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3 3 + 6π₯ 3 4 5 + 3π₯ 5 2π₯ 1+2π₯ 4+3π₯ 5 Warm Up Simplify the fraction:
a) ππ π b) π+ππ π c) π+ππ π π₯ 3 1+2π₯ π₯ 5 4+3π₯ 5 2π₯
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Solve using the quadratic formula
π π π +ππβπ=π π₯= β(6)Β± (6) 2 β4(3)(β4) 2(3) π₯= β6Β± π₯= β π₯= β6β9.17 6 π₯=.52 π₯=2.53
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Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra states that: An nth degree polynomial will have n roots. Example: π¦= π₯ 2 +3π₯+2 is a second degree polynomial, therefore it has 2 roots. Verify this by graphing it on your calculator.
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Fundamental Theorem of Algebra
State the number of roots for each polynomial. Verify by graphing on your calculator. π¦=4π₯β5 π¦=β π₯ 3 β2 π₯ 2 +29π₯β30 π¦= π₯ 4 β2π₯β2 What's the problem with equation number 3? 1st degree, 1 x-intercept 3rd degree, 3 x-intercepts
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Imaginary Numbers What is β1 ? Thatβs a lie. β1 =π
In the 1500s mathematician Rafael Bombelli invented the concept of π so that he could work with β1 .
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Imaginary Numbers We can rewrite the square root of a negative number using π. Example: β16 = 16 β β1 β16 =4π Example 2: β64 =8π
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Imaginary Numbers If π= β1 , what is π 2 ? π 2 = β1 2 π 2 =β1
π 2 = β1 2 π 2 =β1 Simplify: 8π 10π 8π 10π =80 π 2 β80 Simplify: 6β2π+8+4π 14+2π
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Imaginary Numbers Rewrite the expressions below: β36 β81 6π 100
10π β3π 4π 2 5π β π 2 6 3β2π +4π 6π 9π 10 30 β80π β45π 18β8π
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BREAK
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Complex Roots π₯= βπΒ± π 2 β4ππ 2π
Use the quadratic formula to solve: 0= π₯ 2 β4π₯+5 π₯= β(β4)Β± (β4) 2 β4(1)(5) 2(1) π₯= 4Β± β4 2 π₯= 4Β±2π 2 π₯=2Β±π π₯= βπΒ± π 2 β4ππ 2π
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Complex Roots A real number plus or minus an imaginary number is called a complex number. π¦= π₯ 2 β4π₯+5 Has two complex roots: 2+π and 2βπ Graph the function on your calculator. How many times does it cross the x-axis? Since π¦= π₯ 2 β4π₯+5 does not cross the x-axis, it does not have any real roots.
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π ππππ πππππ + π πππππππ πππππ = πππ π
πππππ
Complex Roots Letβs revisit π¦= π₯ 4 β2π₯β2 How many times does the function cross the x-axis? Based on the Fundamental Theorem of Algebra, how many roots should π¦= π₯ 4 β2π₯β2 have? π¦= π₯ 4 β2π₯β2 has 2 real roots and 2 imaginary roots π ππππ πππππ + π πππππππ πππππ = πππ π
πππππ
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Complex Roots π¦= π₯ 2 β4 π¦= π₯ 3 +2 π¦=(π₯β1)(π₯+3)(π₯β6) 2 real
Use your calculator and the Fundamental Theorem of Algebra to determine the number of real and imaginary roots for each polynomial. π¦= π₯ 2 β4 π¦= π₯ 3 +2 π¦=(π₯β1)(π₯+3)(π₯β6) π¦=β3 π₯ 4 +2 π₯ 2 β6 2 real 1 real, 2 complex 3 real 4 complex
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Textbook: 9-90, 9-91, 9-105
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