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Day 146 β Solve π₯ 2 =π
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Finding Square Roots Finding the exact solutions for 0=β16 π‘ involves finding square roots. Every positive number has a positive and a negative square root. The positive square root of 9 is 3. 9 =3 The negative square root of 9 β3. β 9 =β3 When solving a quadratic equation, you can use the symbol Β±, which is read as βplus or minus.β
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Remember to consider both positive and negative square root of 9.
Finding Square Roots Therefore, to solve the equation π₯ 2 =9, take the square root of the expression on each side of the equal sign. Remember to consider both positive and negative square root of 9.
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Example 1 Show that 4 9 = Simplify the square root under the radical signs. 4 9 = 2 3 πππππ’π π 2 3 β 2 3 = = 2 3 πππππ’π π 2β2 3β3 = 4 9 Since 4 9 = 2 3 πππ 4 9 = 2 3 , 4 9 = 4 9 by subsitution
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Example 2 a) π₯ 2 = 4 9 b) π₯ 2 =1.44 c) π₯ 2 =10 Answer a)There are two solutions: 4 9 = 2 3 and β 4 9 =β 2 3 b) There are two solutions: 1.44 =1.2 πππ β 1.44 =β1.2
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Example 2 c) The solutions are 10 πππ β 10 . There is no rational number answer. An approximation can be found by using the key on your calculator. The approximate solutions are 3.16 and β3.16. The results of Example 2 lead to a generalization for solving a quadratic equation of the form π₯ 2 =π. = 10
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Example 2 SOLVING π π =π WHEN πβ₯π If π₯ 2 =π, and πβ₯0, then 1. π₯=Β± π and 2. The solutions are π and β π . For example, if π₯ 2 =16, then π₯=Β± 16 , or Β±4. the solutions are 4 and β4.
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Example 3 Solve each equation. a) πβ2 2 β9=0 b) π₯β2 2 =11
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Answer a) In this equation, the expression πβ2 plays the role of π₯ in the statement π₯ 2 =π. The solutions are 2+3, or 5, and 2β3, or β1. Check each solution in the original equation.
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Answer b) The approximate solutions are , or 5.32, and 2β 11 , or β1.32. Check each solution in the original equation. The solution to Exampl e 3, Part a) will help you sketch the graph of the function π π₯ = π₯β2 2 β9
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Answer axis of symmetry In the form π π₯ = π₯β2 2 β9, the vertex is 2, β9 and the axis of symmetry is π₯=2. Since the coefficient of the quadratic term is positive, the parabola opens upward and has a minimum value. The solutions to π₯β2 2 β9= 0 are 5 and β1. Thus, the graph is a parabola that crosses the x-axis at 5, 0 and (β1, 0). vertex
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