Welcome! Grab a set of interactive notes Section 5: Quadratic Functions Part 1 Topics 7, 9-10 Topics 7 Welcome! Grab a set of interactive notes Homework Assignment: Khan Academy Study Guide Topics 9-10

Section 5: Quadratic Functions Part 1 Topics 7, 9-10 You will: Solve quadratic equations by completing the square Solve quadratic equations by the quadratic formula Recognize when the quadratic formula gives complex solutions.

Objective Solve quadratic equations by completing the square. Section 5: Quadratic Functions Part 1 Topics 7, 9-10 Objective Solve quadratic equations by completing the square.

Section 5: Quadratic Functions Completing the Square In the previous lesson, you solved quadratic equations by isolating x2 and then using square roots. This method works if the quadratic equation, when written in standard form, is a perfect square. When a trinomial is a perfect square, there is a relationship between the coefficient of the x-term and the constant term. X2 + 6x + 9 x2 – 8x + 16 Divide the coefficient of the x-term by 2, then square the result to get the constant term.

Section 5: Quadratic Functions Completing the Square An expression in the form x2 + bx is not a perfect square. However, you can use the relationship shown above to add a term to x2 + bx to form a trinomial that is a perfect square. This is called completing the square.

Section 5: Quadratic Functions Completing the Square Complete the square to form a perfect square trinomial. A. x2 + 2x + A. x2 + 2x + x2 + 2x x2 + –6x Identify b. . x2 + 2x + 1 x2 – 6x + 9

Section 5: Quadratic Functions Completing the Square Complete the square to form a perfect square trinomial. c. x2 + 12x + d. x2 – 5x + x2 + 12x x2 + –5x Identify b. . x2 – 5x + x2 + 12x + 36

Section 5: Quadratic Functions Completing the Square Complete the square to form a perfect square trinomial. e. 8x + x2 + x2 + 8x Identify b. . x2 + 8x + 16

Section 5: Quadratic Functions Completing the Square To solve a quadratic equation in the form x2 + bx = c, first complete the square of x2 + bx. Then you can solve using square roots.

Section 5: Quadratic Functions Completing the Square Solving a Quadratic Equation by Completing the Square

Section 5: Quadratic Functions Completing the Square Solve by completing the square. Check your answer. x2 + 16x = –15 The equation is in the form x2 + bx = c. Step 1 x2 + 16x = –15 Step 2 . Step 3 x2 + 16x + 64 = –15 + 64 Complete the square. Step 4 (x + 8)2 = 49 Factor and simplify. Take the square root of both sides. Step 5 x + 8 = ± 7 Step 6 x + 8 = 7 or x + 8 = –7 x = –1 or x = –15 Write and solve two equations.

  Section 5: Quadratic Functions Completing the Square Solve by completing the square. Check your answer. x2 + 16x = –15 The solutions are –1 and –15. Check x2 + 16x = –15 (–1)2 + 16(–1) –15 1 – 16 –15 –15 –15  x2 + 16x = –15 (–15)2 + 16(–15) –15 225 – 240 –15 –15 –15 

Section 5: Quadratic Functions Completing the Square Solve by completing the square. Check your answer. x2 – 4x – 6 = 0 Write in the form x2 + bx = c. Step 1 x2 + (–4x) = 6 Step 2 . Step 3 x2 – 4x + 4 = 6 + 4 Complete the square. Step 4 (x – 2)2 = 10 Factor and simplify. Take the square root of both sides. Step 5 x – 2 = ± √10 Step 6 x – 2 = √10 or x – 2 = –√10 x = 2 + √10 or x = 2 – √10 Write and solve two equations.

Section 5: Quadratic Functions Completing the Square Solve by completing the square. Check your answer. x2 + 10x = –9 The equation is in the form x2 + bx = c. Step 1 x2 + 10x = –9 Step 2 . Step 3 x2 + 10x + 25 = –9 + 25 Complete the square. Factor and simplify. Step 4 (x + 5)2 = 16 Take the square root of both sides. Step 5 x + 5 = ± 4 Step 6 x + 5 = 4 or x + 5 = –4 x = –1 or x = –9 Write and solve two equations.

Section 5: Quadratic Functions Completing the Square Solve by completing the square. Check your answer. t2 – 8t – 5 = 0 Write in the form x2 + bx = c. Step 1 t2 + (–8t) = 5 Step 2 . Step 3 t2 – 8t + 16 = 5 + 16 Complete the square. Factor and simplify. Step 4 (t – 4)2 = 21 Take the square root of both sides. Step 5 t – 4 = ± √21 Step 6 t = 4 + √21 or t = 4 – √21 Write and solve two equations.

Section 5: Quadratic Functions Completing the Square Solve by completing the square. –3x2 + 12x – 15 = 0 Step 4 (x – 2)2 = –1 Factor and simplify. There is no real number whose square is negative, so there are no real solutions.

Section 5: Quadratic Functions Completing the Square Solve by completing the square. 4t2 – 4t + 9 = 0 Step 1 Divide by 4 to make a = 1. Write in the form x2 + bx = c.

Section 5: Quadratic Functions Completing the Square Solve by completing the square. 4t2 – 4t + 9 = 0 . Step 2 Step 3 Complete the square. Step 4 Factor and simplify. There is no real number whose square is negative, so there are no real solutions.

Objectives Solve quadratic equations by using the Quadratic Formula. Section 5: Quadratic Formula and Discriminant Objectives Solve quadratic equations by using the Quadratic Formula. Determine the number of solutions of a quadratic equation by using the discriminant.

Section 5: Quadratic Formula and Discriminant The Quadratic Formula is the only method that can be used to solve any quadratic equation.

Section 5: Quadratic Formula and Discriminant

Section 5: Quadratic Formula and Discriminant Solve using the Quadratic Formula. 6x2 + 5x – 4 = 0 6x2 + 5x + (–4) = 0 Identify a, b, and c. Use the Quadratic Formula. Substitute 6 for a, 5 for b, and –4 for c. Simplify.

Section 5: Quadratic Formula and Discriminant Solve using the Quadratic Formula. 6x2 + 5x – 4 = 0 Simplify. Write as two equations. Solve each equation.

Section 5: Quadratic Formula and Discriminant Solve using the Quadratic Formula. x2 = x + 20 Write in standard form. Identify a, b, and c. 1x2 + (–1x) + (–20) = 0 Use the quadratic formula. Substitute 1 for a, –1 for b, and –20 for c. Simplify.

Section 5: Quadratic Formula and Discriminant Solve using the Quadratic Formula. x2 = x + 20 Simplify. Write as two equations. x = 5 or x = –4 Solve each equation.

Section 5: Quadratic Formula and Discriminant Solve using the Quadratic Formula. –3x2 + 5x + 2 = 0 –3x2 + 5x + 2 = 0 Identify a, b, and c. Use the Quadratic Formula. Substitute –3 for a, 5 for b, and 2 for c. Simplify.

Section 5: Quadratic Formula and Discriminant Solve using the Quadratic Formula. –3x2 + 5x + 2 = 0 Simplify. Write as two equations. x = – or x = 2 Solve each equation.

Section 5: Quadratic Formula and Discriminant Solve using the Quadratic Formula. 2 – 5x2 = –9x (–5)x2 + 9x + (2) = 0 Write in standard form. Identify a, b, and c. Use the Quadratic Formula. Substitute –5 for a, 9 for b, and 2 for c. Simplify.

Section 5: Quadratic Formula and Discriminant Solve using the Quadratic Formula. 2 – 5x2 = –9x Simplify. Write as two equations. x = – or x = 2 Solve each equation.

Section 5: Quadratic Formula and Discriminant Many quadratic equations can be solved by graphing, factoring, taking the square root, or completing the square. Some cannot be solved by any of these methods, but you can always use the Quadratic Formula to solve any quadratic equation.

Section 5: Quadratic Formula and Discriminant If the quadratic equation is in standard form, the discriminant of a quadratic equation is b2 – 4ac, the part of the equation under the radical sign. Recall that quadratic equations can have two, one, or no real solutions. You can determine the number of solutions of a quadratic equation by evaluating its discriminant.

Section 5: Quadratic Formula and Discriminant

Section 5: Quadratic Formula and Discriminant In Short:

Section 5: Quadratic Formula and Discriminant Find the number of solutions of each equation using the discriminant. A. B. C. 3x2 – 2x + 2 = 0 2x2 + 11x + 12 = 0 x2 + 8x + 16 = 0 a = 3, b = –2, c = 2 a = 2, b = 11, c = 12 a = 1, b = 8, c = 16 b2 – 4ac b2 – 4ac b2 – 4ac (–2)2 – 4(3)(2) 112 – 4(2)(12) 82 – 4(1)(16) 4 – 24 121 – 96 64 – 64 –20 25 b2 – 4ac is negative. There are no real solutions. b2 – 4ac is positive. There are two real solutions. b2 – 4ac is zero. There is one real solution.

Section 5: Quadratic Formula and Discriminant Find the number of solutions of each equation using the discdriminant. D. F. E. 2x2 – 2x + 3 = 0 x2 + 4x + 4 = 0 x2 – 9x + 4 = 0 a = 2, b = –2, c = 3 a = 1, b = 4, c = 4 a = 1, b = –9 , c = 4 b2 – 4ac b2 – 4ac b2 – 4ac (–2)2 – 4(2)(3) 42 – 4(1)(4) (–9)2 – 4(1)(4) 4 – 24 16 – 16 81 – 16 –20 65 b2 – 4ac is negative. There are no real solutions. b2 – 4ac is zero. There is one real solution. b2 – 4ac is positive. There are two real solutions.

Section 5: Quadratic Formula and Discriminant

Section 5: Identifying Quadratic Functions You have identified linear functions by finding that a constant change in x corresponded to a constant change in y. The differences between y-values for a constant change in x-values are called first differences.

Section 5: Identifying Quadratic Functions Notice that the quadratic function y = x2 doe not have constant first differences. It has constant second differences. This is true for all quadratic functions.

Section 5: Identifying Quadratic Functions Tell whether the function is quadratic. Explain. Since you are given a table of ordered pairs with a constant change in x-values, see if the second differences are constant. x y –2 –9 +1 +7 +1 –6 +0 +6 –1 –2 –1 1 Find the first differences, then find the second differences. 2 7 The function is not quadratic. The second differences are not constant.

Section 5: Quadratic Functions Part 1 Topics Homework Khan Academy Study Guide