 # Algebra 1B Chapter 9 Solving Quadratic Equations The Discriminant.

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Algebra 1B Chapter 9 Solving Quadratic Equations The Discriminant

Warm Up Use the Quadratic Formula to solve each equation. 1. x 2 – 5x – 6 = 0 2. 2x 2 + 2x – 24 = 0 3. x 2 + 10x + 25 = 0 0, –5 3, –4 1, –6

discriminant Vocabulary

If the quadratic equation is in standard form, its discriminant is b 2 – 4ac. Notice that this is the expression under the square root in the Quadratic Formula. 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 the discriminant.

Additional Example 1A: Using the Discriminant Find the number of solutions of 3x 2 – 2x + 2 = 0 by using the discriminant. a = 3, b = –2, c = 2 b 2 – 4ac =(–2) 2 – 4(3)(2) = 4 – 24 = –20 Identify the values of a, b, and c. Substitute 3, –2, and 2 for a, b, and c. Simplify. b 2 – 4ac is negative. There are no real solutions.

Additional Example 1B: Using the Discriminant Find the number of solutions of 2x 2 + 11x + 12 = 0 by using the discriminant. a = 2, b = 11, c = 12 b 2 – 4ac =11 2 – 4(2)(12) = 121 – 96 = 25 Identify the values of a, b, and c. Substitute 2, 11, and 12 for a, b, and c. Simplify. b 2 – 4ac is positive. There are two solutions.

Additional Example 1C: Using the Discriminant Find the number of solutions of x 2 + 8x + 16 = 0 by using the discriminant. a = 1, b = 8, c = 16 b 2 – 4ac =8 2 – 4(1)(16) = 64 – 64 = 0 Identify the values of a, b, and c. Substitute 1, 8, and 16 for a, b, and c. Simplify. b 2 – 4ac is zero. There is one solution.

In Your Notes! Example 1a Find the number of solutions of 2x 2 – 2x + 3 = 0 using the discriminant. a = 2, b = –2, c = 3 b 2 – 4ac =(–2) 2 – 4(2)(3) = 4 – 24 = –20 Identify the values of a, b, and c. Substitute 2, –2, and 3 for a, b, and c. Simplify. b 2 – 4ac is negative. There are no real solutions.

In Your Notes! Example 1b Find the number of solutions of x 2 + 4x + 4 = 0 using the discriminant. a = 1, b = 4, c = 4 b 2 – 4ac =4 2 – 4(1)(4) = 16 – 16 = 0 Identify the values of a, b, and c. Substitute 1, 4, and 4 for a, b, and c. Simplify. b 2 – 4ac is zero. There is one real solution.

Remember! The solutions to a quadratic are the same as the x-intercepts of the related function. The discriminant can be used to find the number of x-intercepts.

Additional Example 2A: Using the Discriminant to Find the number of x-Intercepts Find the number of x-intercepts of y = 2x 2 – 9x + 5 using the discriminant. a = 2, b = –9, c = 5 Identify the values of a, b, and c. b 2 – 4ac =(–9) 2 – 4(2)(5) = 81 – 40 = 41 Simplify. b 2 – 4ac is positive. Therefore, the function y = 2x 2 – 9x + 5 has two x- intercepts. The graph intercepts the x-axis twice. Substitute 2,–9, and 5 for a, b, and c.

Additional Example 2B: Using the Discriminant to find the number of x-Intercepts Find the number of x-intercepts of y = 6x 2 – 4x + 5 using the discriminant. a = 6, b = –4, c = 5 Identify the values of a, b, and c. b 2 – 4ac =(–4) 2 – 4(6)(5) = 16 – 120 = –104 Simplify. b 2 – 4ac is negative. Therefore, the function y = 6x 2 – 4x + 5 has no x- intercepts. The graph does not intercept the x-axis. Substitute 6, –4, and 5 for a, b, and c.

In Your Notes! Example 2a Find the number of x-intercepts of y = 5x 2 + 3x + 1 by using the discriminant. a = 5, b = 3, c = 1 Identify the values of a, b, and c. b 2 – 4ac =3 2 – 4(5)(1) = 9 – 20 = –11 Simplify. b 2 – 4ac is negative. Therefore, the function y = 5x 2 + 3x + 1 has no x- intercepts. The graph does not intercept the x-axis. Substitute 5, 3, and 1 for a, b, and c.

In Your Notes! Example 2b Find the number of x-intercepts of y = x 2 – 9x + 4 by using the discriminant. a = 1, b = –9, c = 4 Identify the values of a, b, and c. b 2 – 4ac =(–9) 2 – 4(1)(4) = 81 – 16 = 65 Simplify. b 2 – 4ac is positive. Therefore, the function y = x 2 – 9x + 4 has two x- intercepts. The graph intercepts the x-axis twice. Substitute 1, –9, and 4 for a, b, and c.

The ringer on a carnival strength test is 2 feet off the ground and is shot upward with an initial velocity of 30 feet per second. Will it reach a height of 20 feet? Use the discriminant to explain your answer. Additional Example 3: Physical Science Application The height h in feet of an object shot straight up with initial velocity v in feet per second is given by h = –16t 2 + vt + c, where c is the initial height of the object above the ground.

Additional Example 3 Continued h = –16t 2 + vt + c 20 = –16t 2 + 30t + 2 0 = –16t 2 + 30t + (–18) b 2 – 4ac 30 2 – 4(–16)(–18) = –252 Substitute 20 for h, 30 for v, and 2 for c. Subtract 20 from both sides. Evaluate the discriminant. Substitute – 16 for a, 30 for b, and – 18 for c. The discriminant is negative, so there are no real solutions. The ringer will not reach a height of 20 feet.

If the object is shot straight up from the ground, the initial height of the object above the ground equals 0. Helpful Hint

In Your Notes! Example 4 What if…? Suppose a weight is shot straight up from the ground with an initial velocity of 20 feet per second. Will it reach the height of 45 feet? Use the discriminant to explain your answer. h = –16t 2 + vt + c 45 = –16t 2 + 20t Substitute 45 for h, and 20 for v. 0 = –16t 2 + 20t + (–45) Subtract 45 from both sides. b 2 – 4ac 20 2 – 4(–16)(–45) = –2080 Evaluate the discriminant. Substitute – 16 for a, 20 for b, and – 45 for c. No; for the equation 45 = –16t 2 + 20t, the discriminant is negative, so the weight will not ring the bell.