Chapter 10 Section 3 Solving Quadratic Equations by the Quadratic Formula

Learning Objectives Solve quadratic equation be the quadratic formula Determine the number of solutions to a quadratic equation using the discriminant

Key Vocabulary Quadratic formula Discriminant

Quadratic Formula Standard form of a quadratic equation is ax 2 + bx + c = 0 where a is the coefficient of the squared term and b is the coefficient of the first-degree term and c is the constant. It is important to label the a, b, and c with the correct sign when substituting into the quadratic formula.

Solve the Quadratic Equation by Factoring x 2 + 7x + 10 = 0 (x + 2)(x + 5) = 0 x + 2 = 0 and x + 5 = 0 x = -2 and x = -5 Not all equations can be solve by factoring.

Solve the Quadratic Equation by Completing the Square x 2 + 7x + 10 = 0 ( ½ )(7) = 3.5 (3.5) 2 = x 2 + 7x = -10 x 2 + 7x = x 2 + 7x = 2.25 (x + 3.5) 2 = 2.25 x = ±1.5 x = -3.5 ±1.5 x = and x = -3.5 – 1.5 x = -2 and x = -5 Perfect Square Trinomial

To Solve a Quadratic Equation by the Quadratic Formula 1.Write the equation in standard form to determine the values of a, b, and c. 2.Substitute the values for a, b and c from the equation in standard form into the quadratic formula and evaluate. 3.Check by placing back into original equation.

Solve using the Quadratic Formula x 2 + 7x + 10 = 0 Already in standard form a = 1 b = 7 c = 10 x = -2 x = -5

Solve using the Quadratic Formula 18x 2 – 3x – 1 = 0 Already in standard form a = 18 b = -3 c = -1

Solve using the Quadratic Formula 5n 2 + 2n - 1 = 0 Already in standard form a = 5 b = 2 c = -1

x 2 = 4x – 1 Put in standard form x 2 – 4x + 1 = 0 a = 1 b = -4 c = 1 Solve using the Quadratic Formula

4x 2 = 2x – 3 Put in standard form 4x 2 – 2x + 3 = 0 a = 4 b = -2 c = 3 Solve using the Quadratic Formula < 0

m 2 = 64 Put in standard form m 2 +0x – 64 = 0 a = 1 b = 0 c = -64 Solve using the Quadratic Formula

x 2 +14x +45 = 0 Already in standard form a = 1 b = 14 c = 45 x = -5 x = -9 Solve using the Quadratic Formula

12x 2 - 4x - 1 = 0 Already in standard form a = 12 b = -4 c = -1 Solve using the Quadratic Formula

3x 2 + 4x - 8 = 0 Already in standard form a = 3 b = 4 c = -8 Solve using the Quadratic Formula

x 2 = 8x – 6 Put in standard form x 2 – 8x + 6 = 0 a = 1 b = -8 c = 6 Solve using the Quadratic Formula

a 2 – 121 = 0 Put in standard form a 2 + 0x – 121 = 0 a = 1 b = 0 c = -121 Solve using the Quadratic Formula

4x 2 - 2x + 3 = 0 Already in standard form a = 4 b = -2 c = 3 No real solution Solve using the Quadratic Formula < 0

The length of a rectangle is 1 ft more than three times the width. Find the dimensions of the rectangle if the area is 30 ft 2, find the length and width A = LW Let x = width length = 3x + 1 a = 3 (3x + 1)x = 30b = 1 c = -30 3x 2 + x = 30 put in standard form 3x 2 + x – 30 = 0 W = 3 L = (3)(3) + 1 = 10 Solve an Application Problem using the Quadratic Formula

Determine the Number of Solutions to a Quadratic Equation Using the Discriminant The expression under the radical is called the discriminant and can be used to determine the number of solutions b 2 – 4ac is called the discriminant b 2 – 4ac > 0 two distinct real number solutions b 2 – 4ac = 0one real number solution b 2 – 4ac < 0no real number solution, negative

Determine the Number of Solutions to a Quadratic Equation Using the Discriminant 2x 2 + 3x = 5b 2 – 4ac (3) 2 – (4)(2)(-5) Put in standard form9 – (8)(-5) x 2 + 3x – 5 = 049 a = 2 49 > 0 b = 3 c = -5Two distinct real number solutions

Determine the Number of Solutions to a Quadratic Equation Using the Discriminant 2x 2 + x + 3 = 0b 2 – 4ac (1) 2 – (4)(2)(3) Already in standard form1 – (8)(3) 1 – 24 a = b = 1 c = 3-23 < 0 No real number solutions

Determine the Number of Solutions to a Quadratic Equation Using the Discriminant 9x x + 16 = 0b 2 – 4ac (-24) 2 – (4)(9)(16) Already in standard form256 – (36)(16) 576 – 576 a = 9 0 b = -24 c = 160 = 0 One real number solutions

Determine the Number of Solutions to a Quadratic Equation Using the Discriminant 2x x - 7 = 0b 2 – 4ac (-3) 2 – (4)(2)(-7) Already in standard form9 – (8)(-7) a = 2 65 b = -3 c = -765 > 0 Two distinct real number solutions

Determine the Number of Solutions to a Quadratic Equation Using the Discriminant 3x 2 = - x + 4b 2 – 4ac (1) 2 – (4)(3)(-4) Put in standard form1 – (12)(-4) 1 – 48 3x 2 + x – 4 = 0-47 a = < 0 b = 1 c = -4No real number solutions

Remember The equation should be in standard form so that you can correctly determine a, b, and c. Make sure that you label a, b, and c with the correct sign. Check your answers by placing them back into the original equation. Use the discriminant to determine the number of solutions

HOMEWORK 10.3 Page 609 – 610: # 27, 29, 35, 36, 43, 47, 51, 59