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9.3 Solve Using Square Roots
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What We Will Learn Solve quadratics by using square roots
Story problems
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Essential question How can you determine the number of solutions of a quadratic equation of the form ππ₯ 2 +π=0?
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Ex. 1 Using Square Roots 3π₯ 2 β27=0 +27+27 3π₯ 2 =27 3π₯ 2 3 = 27 3
3π₯ 2 3 = 27 3 π₯ 2 =9 π₯ 2 = 9 π₯=Β±3 Two solutions π₯ 2 β10=β10 π₯ 2 =0 π₯ 2 = 0 π₯=0 One solution β5π₯ 2 +11=16 β11 β11 β5π₯ 2 =5 β5π₯ 2 β5 = 5 β5 π₯ 2 =β1 π₯ 2 = β1 No solution
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Your Practice β3π₯ 2 =β75 π₯ 2 +12=0 β12 β12 4π₯ 2 β15=β15 +15 +15
β3π₯ 2 β3 = β75 β3 π₯ 2 =25 π₯ 2 = 25 π₯=Β±5 π₯ 2 +12=0 β12 β12 π₯ 2 =β12 π₯ 2 = β12 No solution 4π₯ 2 β15=β15 4π₯ 2 =0 4π₯ 2 4 = 0 4 π₯ 2 =0 π₯ 2 = 0 π₯=0
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Ex. 2 solving 2 π₯β1 2 =25 π₯β1 2 = 25 π₯β1=Β±5 π₯β1=5, π₯β1=β5 +1 +1, +1 +1
π₯β1 2 =25 π₯β1 2 = 25 π₯β1=Β±5 π₯β1=5, π₯β1=β5 +1 +1, π₯=6, π₯=β4 Steps 1. Get by itself both sides 3. solve for letter 4 π₯β3 2 =36 4 π₯β = 36 4 π₯β3 2 =9 π₯β3 2 = 9 π₯β3=Β±3 π₯β3=3, π₯β3=β3 π₯=6, π₯=0
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Your Practice 9 π₯+1 2 +5=149 β5 β5 9 π₯+1 2 =144 9 π₯+1 2 9 = 144 9
9 π₯ =149 β β5 9 π₯+1 2 =144 9 π₯ = 144 9 π₯+1 2 =16 π₯ = 16 π₯+1=Β±4 π₯+1=4, π₯+1=β4 β1 β β1 β1 π₯=3, π₯=β5
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Ex. 3 Approximating Square Roots
Solve and Round to nearest hundredth 4π₯ 2 β13=15 4π₯ 2 =28 4π₯ 2 4 = 28 4 π₯ 2 =7 π₯ 2 = 7 π₯=Β±2.65
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Ex. 4 Story Problem A touch tank has a height of 3 feet. Its length is three times its width. The volume of the tank is 270 cubic feet. Find the length and width of the tank. Need V = lwh V = volume l = length w = width h = height V = 270 h = 3 l = 3w w = w 270 = 3w(w)(3) 270= 9π€ 2 270 9 = 9π€ 2 9 30= π€ 2 30 = π€ 2 5.5=π€ l = 3(5.5) l = 16.5
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