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Section 8.6-8.7 Day 3 Solving π π₯ 2 +ππ₯+π=0
Algebra 1
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Learning Targets Factor trinomials in the form π π₯ 2 +ππ₯+π
Solve algebraic equations of the form π π₯ 2 +ππ₯+π=0
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Multiple Variables Example 1
Factor π₯ 2 β14π₯π¦β51 π¦ 2 πβπ=β51 π¦ 2 π=β14π¦ Answer: (π₯+3π¦)(π₯β17π¦)
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Multiple Variables Example 2
Factor π 2 +10ππβ39 π 2 πβπ=β39 π 2 π=10π Answer: (π+13π)(πβ3π)
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2 Step Factoring β Level 1 Example 3
GCF: 4 4( π 2 +8πβ48) Answer: 4(πβ4)(π+12)
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2 Step Factoring: Level 1 Example 4
GCF: 2 2(β4 π₯ 2 +19π₯+30) Answer: 2(4π₯+5)(βπ₯+6)
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2 step Factoring: Level 2 Example 6
GCF: 4π₯ 4π₯(3 π₯ 2 β17π₯+20) Answer: 4π₯(3π₯β5)(π₯β4)
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Solving π π₯ 2 +ππ₯+π=0 Example 5
Solve 6 π₯ 3 β51 π₯ 2 +90π₯=0 3π₯ 2 π₯ 2 β17π₯+30 =0 3π₯ π₯β6 2π₯β5 =0 Answer: π₯=0, π₯=6, π₯= 5 2
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Solving π π₯ 2 +ππ₯+π=0 Example 7
Solve 3 π 5 β6 π 4 =105 π 3 Rewrite: 3 π 5 β6 π 4 β105 π 3 =0 GCF: 3 π 3 3 π 3 π 2 β2πβ35 =0 3 π 3 πβ7 π+5 =0 Answer: π=0, π=7, π=β5
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Solving π π₯ 2 +ππ₯+π=0 Example 8 (PG 512: #4)
A person throws a ball upward from a 506-foot tall building. The ballβs height β in feet after π‘ seconds is given by the equation β= β16 π‘ 2 +48π‘+506. The ball lands on a balcony that is 218 feet above the ground. How many seconds was it in the air? 218=β16 t 2 +48t+506 0=β16 π‘ 2 +48π‘+288 0=β16 π‘ 2 β3π‘+18 =β16(π‘β6)(π‘+3) π‘=6 π ππππππ
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Problem 1 Factor π₯ 2 +9π₯+20 Answer: (π₯+5)(π₯+4)
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Problem 2 Solve 2 π₯ 2 +22π₯=β20 Answer: 2 π₯+10 (π₯+1) π₯=β10 and π₯=β1
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Problem 3 Solve 2 π₯ 2 β3π₯=20 Answer: π₯=4, π₯=β 5 2
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