5 6 The Quadratic Formula and the Discriminant

5. 6 The Quadratic Formula and the Discriminant Algebra 2

Learning Target l Solve equations using the quadratic formula. l Use the discriminant to determine the nature of the roots of a quadratic equation.

Derive the quadratic formula from ax 2 + bx + c = 0 a≠ 0 General form of a quadratic equation. Divide all by a Simplify Subtract c/a on both sides. Multiply by ½ and square the result.

Derive the quadratic formula from ax 2 + bx + c = 0 a≠ 0 Add the result to both sides. Simplify Multiply by common denominator Simplify

Derive the quadratic formula from ax 2 + bx + c = 0 a≠ 0 Square root both sides Simplify Common denominator/subtract from both sides Simplify

Quadratic Formula l The solutions of a quadratic equation of the form ax 2 + bx + c with a ≠ 0 are given by this formula: MEMORIZE!!!!

Ex. 1: Solve t 2 – 3 t – 28 = 0 a = 1 b = -3 c = -28 There are 2 distinct roots—Real and rational.

Ex. 1: Solve t 2 – 3 t – 28 = 0 CHECK: t 2 – 3 t – 28 = 0 72 – 3(7) – 28 = 0 49 – 21 – 28 = 0 49 – 49 = 0 CHECK: t 2 – 3 t – 28 = 0 (-4)2 – 3(-4) – 28 = 0 16 + 12 – 28 = 0 28 – 28 = 0

Ex. 1: Solve t 2 – 3 t – 28 = 0 -- GRAPH

Ex. 2: Solve x 2 – 8 x + 16 = 0 a = 1 b = -8 c = 16 There is 1 distinct root—Real and rational.

Ex. 2: Solve x 2 – 8 x + 16 = 0 CHECK: x 2 – 8 x + 16 = 0 (4)2 – 8(4) + 16 = 0 16 – 32 + 16 = 0 32 – 32 = 0 There is 1 distinct root—Real and rational.

Ex. 2: Solve x 2 – 8 x + 16 = 0 -GRAPH

Ex. 3: Solve 3 p 2 – 5 p + 9 = 0 a = 3 b = -5 c = 9 There is 2 imaginary roots.

Ex. 3: Solve 3 p 2 – 5 p + 9 = 0 NOTICE THAT THE PARABOLA DOES NOT TOUCH THE X-AXIS.

Note: l These three examples demonstrate a pattern that is useful in determining the nature of the root of a quadratic equation. In the quadratic formula, the expression under the radical sign, b 2 – 4 ac is called the discriminant. The discriminant tells the nature of the roots of a quadratic equation.

DISCRIMINANT l The discriminant will tell you about the nature of the roots of a quadratic equation. Equation t 2 – 3 t – 28 = 0 Value of the discriminant Roots b 2 – 4 ac = (-3)2 – 4(1)(-28) = 121 {7, - 4} x 2 – 8 x + 16 = 0 b 2 – 4 ac = {0} Nature of roots 2 real roots 1 real root (-8)2 – 4(1)(16) = 0 3 p 2 – 5 p + 9 = 0 b 2 – 4 ac = (-5)2 – 4(3)(9) = -83 2 imaginary roots

Ex. 4: Find the value of the discriminant of each equation and then describe the nature of its roots. 2 x 2 + x – 3 = 0 a = 2 b = 1 c = -3 b 2 – 4 ac = (1)2 – 4(2)(-3) = 1 + 24 = 25 The value of the discriminant is positive and a perfect square, so 2 x 2 + x – 3 = 0 has two real roots and they are rational.

Ex. 5: Find the value of the discriminant of each equation and then describe the nature of its roots. x 2 + 8 = 0 a=1 b=0 c=8 b 2 – 4 ac = (0)2 – 4(1)(8) = 0 – 32 = – 32 The value of the discriminant is negative, so x 2 + 8 = 0 has two imaginary roots.

Pair-share l pp. 295 #16 -45 even