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TODAY IN GEOMETRY⦠Review methods for solving quadratic equations Learning Goal: You will solve quadratics by using the Quadratic Formula In Class Practice
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REVIEW: Compare the different ways to solve the following quadratic equations.
π₯ 2 β11π₯+30= π₯ 2 β7π₯+2=0 2. Coefficient greater than 1 1. Coefficient of 1 β + 12 β + βπ βπ 6 π₯ 2 β3π₯β4π₯+2=0 6 π₯ 2 β3π₯ β4π₯+2 =0 3π₯ π₯β1 β2 π₯β1 =0 3π₯β2 π₯β1 =0 3π₯β2= π₯β1=0 30 β7 βπ βπ β11 π₯β5 π₯β6 =0 π₯β5= π₯β6=0 3π₯=2 π= π π π=π π=π π=π
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Quadratic Formula Rap
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SOLVING QUADRATICS BY THE QUADRATIC FORMULA
If we cannot solve by factoring we can find the zeros by using the QUADRATIC FORMULA: If ππ₯ 2 +ππ₯+π=0 then π= βπΒ± π π βπππ ππ
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SOLVING QUADRATICS BY THE QUADRATIC FORMULA
EXAMPLE: Solve the following quadratics. 1. π₯ 2 +3π₯β2= π₯ 2 β12π₯+9=0 π π π π π π 1 π₯= βπΒ± π 2 β4ππ 2π π₯= β(3)Β± (3) 2 β4(1)(β2) 2(1) π₯= β3Β± 9β(β8) 2 π₯= β3Β± π₯= βπΒ± π 2 β4ππ 2π π₯= β(β12)Β± (β12) 2 β4(4)(9) 2(1) π₯= 12Β± 144β144 2 π₯= 12Β± 0 2 π=π π= βπ+ ππ π π= βπβ ππ π
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SOLVING QUADRATICS BY THE QUADRATIC FORMULA
EXAMPLE: Solve the following quadratics. 1. βπ₯ 2 +4π₯β5= β5π₯ 2 +10π₯β5=0 π π π π π π 1 π₯= βπΒ± π 2 β4ππ 2π π₯= β(4)Β± (4) 2 β4(β1)(β5) 2(β1) π₯= β4Β± 16β(5) β2 π₯= β4Β± 11 β2 π₯= βπΒ± π 2 β4ππ 2π π₯= β(10)Β± (10) 2 β4(β5)(β5) 2(β5) π₯= β10Β± 100β100 β10 π₯= β10Β± 0 β10 π₯= β β2 π₯= β4β 11 β2 π=π π=πβ ππ π π=π+ ππ π
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Solving Quadratics by Quadratic Formula WS
IN CLASS WORK #5: Solving Quadratics by Quadratic Formula WS Previous Assignments: HW#1: Substitution and Elimination WS HW#2: Factor Binomials and Trinomials WS HW#3: Factor Trinomials with coeff. greater than 1 WS HW#4: Solving Quadratics by Factoring WS
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