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**Solving Quadratic Equations Lesson 9-3**

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**Standard form of a quadratic function**

A quadratic equation is a equation that can be written in the form π π π +ππ+π=π Where πβ 0. This form is called the standard form of a quadratic equation

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**Quadratic equation can be solved by a variety of methods**

Graphing Squared Root property Factoring Completing the square Quadratic formula

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**Solutions of a Quadratic Equation**

A quadratic equation can have two, one or no real number solutions The solutions of a quadratic equation are the x- intercepts of the related function The solutions of a quadratic equation are often called Roots of the equation or Zeros of the function

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**Steps in Solving Quadratic Equations**

If the equation is in the form (ax+b)2 = c, use the square root property to solve. If not solved in step 1, write the equation in standard form. Try to solve by factoring. If you havenβt solved it yet, use the quadratic formula.

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**are the x-intercepts 2 and -2**

Solving Quadratic Equation by graphing Graph the parabola by: y x Finding the equation of the axis of symmetry The solution of π₯ 2 β4=0 are the x-intercepts 2 and -2 π₯= βπ 2π x= 0 2(1) π₯=0 Finding the vertex (0) 2 β4=β Vertex = (0,-4) Making a table using the x-values around the axis of symmetry 2 -2 x π π βπ y 2 (π) π βπ -2 (βπ) π βπ Graphing each point on a coordinate plane The roots of the equation or solution are the x-intercept or zeros of the related quadratic function

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**Examples of solving by graphing**

What are the solution of each equation? Use a graph of the related function π₯ 2 β1=0 π₯ 2 =0 π₯ 2 +1=0 There are two solutions 1 and -1 There is one solutions There is no real number solution

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**Solving Quadratic Equation by using the square root**

Square Root Property If b is a real number and a2 = b, then You can solve equations of the form π π =π by finding the square root of each side. For example π₯ 2 =81 π₯=Β± 81 π₯=Β±9

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**Solving Quadratic Equation by using the square root**

Example Solve: π΅) 3π₯ 2 β75=0 A) (y β 3)2 = 4 3π₯ 2 =75 π₯ 2 = 75 3 π¦β3=Β±2 π₯ 2 = 25 π¦=3Β±2 π¦=1 ππ π¦=5 π₯=Β± 25 π₯=Β±5

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**Solving Quadratic Equation by using**

Quadratic formula This method will work to solve ALL quadratic equations For many equations it takes longer than some of the other methods.

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**Solving Quadratic Equation by using**

Quadratic formula A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions. π= βπΒ± π π βπππ ππ When getting your solution, if the radicand in the quadratic formula is not a perfect square, you can use a calculator to approximate the solution

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**Solve x2 + x β = 0 by the quadratic formula.**

Example Solve x2 + x β = 0 by the quadratic formula. x2 + 8x β 20 = (multiply both sides by 8) a = 1, b = 8, c = ο20 = β or β8β12 2 = β20 2 ππ 4 2 π₯=β10 ππ π₯=2

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**So there is no real solution.**

Example Solve x(x + 6) = ο30 by the quadratic formula. (Simplify the polynomial and write it on standard form) x2 + 6x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution. Square root canβt be negative.

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**Solve 2x = x2 - 8. x2 β 2x β 8 = 0 Let a = 1, b = -2, c = -8 Example**

write the equation on standard form x2 β 2x β 8 = Let a = 1, b = -2, c = -8 π₯= ππ π₯= 2β6 2 π₯= ππ π₯=β2

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The Discriminat Before you solve a quadratic equation you can determine how many real-number solutions it has by using discriminant. The expression under the radical sign in the formula (b2 β 4ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.

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Example Use the discriminant to determine the number and type of solutions for the following equation. 5 β 4x + 12x2 = 0 a = 12, b = β4, and c = 5 b2 β 4ac = (β4)2 β 4(12)(5) = 16 β 240 = β224 Because the discriminant is negative, the equation has no-real number solution.

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**Because the discriminant is positive, the equation has two solution.**

Example Use the discriminant to determine the number and type of solutions for the following equation. 6 π₯ 2 β5π₯=7 6 π₯ 2 β5π₯β7=0 a = 6, b = β5, and c = -7 b2 β 4ac = (β6)2 β 4(6)(-7) = = 232 Because the discriminant is positive, the equation has two solution.

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