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Solving Quadratic Equations using the Quadratic Formula
9.5 Solving Quadratic Equations using the Quadratic Formula Students will be able to solve quadratic equations using the Quadratic Formula. Students will be able to interpret the discriminant. Students will be able to choose efficient methods for solving quadratic equations.
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Students will be able to solve quadratic equations using the Quadratic Formula.
By completing the square for the quadratic equation π π₯ 2 +ππ₯+π=0, you can develop a formula that gives the solution of any quadratic equation in standard from. This formula is called the Quadratic Formula.
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Students will be able to solve quadratic equations using the Quadratic Formula.
The real solutions of the quadratic equation π π₯ 2 +ππ₯+π=0 are where πβ 0 and π 2 β4ππβ₯0.
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Students will be able to solve quadratic equations using the Quadratic Formula.
Solve 2 π₯ 2 β5π₯+3=0 using the Quadratic Formula. Remember π=2, π=β5, π=3 Plug into the Quadratic Formula
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Students will be able to solve quadratic equations using the Quadratic Formula.
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Students will be able to solve quadratic equations using the Quadratic Formula.
Solve π₯ 2 β6π₯+5=0 using the Quadratic Formula. Remember π=1, π=β6, π=5 Plug into the Quadratic Formula
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Students will be able to solve quadratic equations using the Quadratic Formula.
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Students will be able to solve quadratic equations using the Quadratic Formula. You Try!!
Solve β3π₯ 2 +2π₯+7=0 using the Quadratic Formula. Remember π=β3, π=2, π=7 Plug into the Quadratic Formula
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Students will be able to solve quadratic equations using the Quadratic Formula.
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2. Students will be able to interpret the discriminant.
The expression π 2 β4ππ in the Quadratic Formula is called the discriminant. Because the discriminant is under the radical symbol, you can use the value of the discriminant to determine the number of real solutions of a quadratic equation and the number of x-intercepts of the graph of the related function. π 2 β4ac>0 Two real solutions Two x-intercepts π 2 β4ac=0 One real solution One x-intercept π 2 β4ac<0 No real solutions No x-intercept
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2. Students will be able to interpret the discriminant.
Determine the number of real solutions of π₯ 2 +8π₯β3=0. π 2 β4ac= 8 2 β4β1β(β3) =64β β12 =64+12 =76 The discriminant is greater than 0. So, the equation has two real solutions.
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2. Students will be able to interpret the discriminant.
Determine the number of real solutions of 9 π₯ 2 +1=6π₯. Write the equation in standard form: 9 π₯ 2 β6π₯+1=0 π 2 β4ac= (β6) 2 β4β9β1 =36β36 =0 The discriminant equals 0. So, the equation has one real solution.
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2. Students will be able to interpret the discriminant.
You Try!! Determine the number of real solutions of 6 π₯ 2 +2π₯=β1. Write the equation in standard form: 6 π₯ 2 +2π₯+1=0 π 2 β4ac= (2) 2 β4β6β1 =4β24 =β20 The discriminant is less than 0. So, the equation has no real solutions.
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Methods for Solving Quadratic Equations.
3. Students will be able to choose efficient methods for solving quadratic equations. Methods for Solving Quadratic Equations. Method Advantages Disadvantages Factoring Straight forward when the equation can be factored easily Some equations are not factorable Using Square Roots Use to solve equations of the form π₯ 2 =π Can only be used for certain equations Completing the Square Best used when π=1 and π is even May involve difficult calculations Quadratic Formula Can be used for any quadratic equation Gives exact solutions Takes time to do calculations
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