Solving Quadratic Equations MCR3U: Functions Recall that to solve a - PDF document

f u n c t i o n s f u n c t i o n s Solving Quadratic Equations MCR3U: Functions Recall that to solve a quadratic equation means to find all values of the independent variable to satisfy a given equation. For example, the quadratic equation x 2 + 5 x + 6 = 0 has two solutions, x = − 2 or x = − 3. Solving Quadratic Equations There are several techniques that we can use to solve quadratic equations. Part 1: Factoring and Quadratic Formula J. Garvin J. Garvin — Solving Quadratic Equations Slide 1/18 Slide 2/18 f u n c t i o n s f u n c t i o n s Solving by Factoring Solving by Factoring Often, the easiest method of solving a quadratic equation is Example to factor it. Solve 10 x 2 + 13 x − 3 = 0. We have reviewed several methods of factoring for: • simple trinomials (inspection) 10 x 2 + 13 x − 3 = 0 • complex trinomials (decomposition) • perfect squares ( b = 2 √ a √ c ) 10 x 2 + 15 x − 2 x − 3 = 0 • differences of squares (inspection) 5 x (2 x + 3) − 1(2 x + 3) = 0 (5 x − 1)(2 x + 3) = 0 • grouping x = 1 5 or − 3 2 J. Garvin — Solving Quadratic Equations J. Garvin — Solving Quadratic Equations Slide 3/18 Slide 4/18 f u n c t i o n s f u n c t i o n s Solving by Factoring Solving by Completing the Square Your Turn Sometimes it is not easy to factor a quadratic equation. √ Solve 45 x 3 − 5 x = 0. The equation x 2 + 4 x + 2 = 0 has solutions − 2 + 2 and √ − 2 − 2. These are not “obvious” in any way. Another technique that can be used to solve quadratic 45 x 3 − 5 x = 0 equations is Completing the Square. 5 x (9 x 2 − 1) = 0 While we typically associate this technique with converting a quadratic from standard to vertex form, this form is easy to 5 x (3 x − 1)(3 x + 1) = 0 isolate the independent variable and, thus, solve for any x = 0 , 1 3 or − 1 intercepts. 3 J. Garvin — Solving Quadratic Equations J. Garvin — Solving Quadratic Equations Slide 5/18 Slide 6/18

f u n c t i o n s f u n c t i o n s Solving by Completing the Square Solving by Completing the Square Example Now that the function is in vertex form, isolate x . Determine the x -intercepts of f ( x ) = 2 x 2 + 12 x + 13. 0 = 2( x + 3) 2 − 5 5 = 2( x + 3) 2 5 2 = ( x + 3) 2 f ( x ) = 2 x 2 + 12 x + 13 = 2( x 2 + 6 x ) + 13 � 5 ± 2 = x + 3 = 2( x 2 + 6 x + 9 − 9) + 13 � 5 = 2( x + 3) 2 − 5 − 3 ± 2 = x � � 5 5 The zeroes of f ( x ) are − 3 + 2 and − 3 − 2 . J. Garvin — Solving Quadratic Equations J. Garvin — Solving Quadratic Equations Slide 7/18 Slide 8/18 f u n c t i o n s f u n c t i o n s Solving by Completing the Square Solving Using the Quadratic Formula Completing the square can be a tedious process, prone to The Quadratic Formula The zeroes of a quadratic function, f ( x ) = ax 2 + bx + c , are errors. √ b 2 − 4 ac For this reason, we develop an explicit formula that can given by x = − b ± . determine the zeroes of a quadratic function in standard 2 a form, f ( x ) = ax 2 + bx + c . The formula itself comes directly from completing the square. ax 2 + bx + c = 0 � x 2 + b � a ax = − c a + b 2 4 a 2 − b 2 � � x 2 + b a = − c 4 a 2 � 2 − b 2 � x + b a 4 a = − c 2 a J. Garvin — Solving Quadratic Equations J. Garvin — Solving Quadratic Equations Slide 9/18 Slide 10/18 f u n c t i o n s f u n c t i o n s Solving Using the Quadratic Formula Solving Using the Quadratic Formula Example Determine the zeroes of g ( x ) = 2 x 2 + 8 x + 3. � 2 = − c + b 2 � x + b a 2 a 4 a � 2 = b 2 − 4 ac � x + b x = − 8 ± √ 8 2 − 4(2)(3) 2 a 4 a 2 2(2) � b 2 − 4 ac = − 8 ±√ 64 − 24 x + b 2 a = ± 4 4 a 2 √ √ = − 8 ± 40 b 2 − 4 ac x = − b 4 2 a ± √ = − 8 ± 2 10 2 a √ 4 b 2 − 4 ac √ x = − b ± = − 4 ± 10 2 2 a √ √ The zeroes of g ( x ) are x = − 4+ 10 and x = − 4 − 10 . 2 2 J. Garvin — Solving Quadratic Equations J. Garvin — Solving Quadratic Equations Slide 11/18 Slide 12/18

f u n c t i o n s f u n c t i o n s Solving Using the Quadratic Formula Solving Using the Quadratic Formula Your Turn Example Determine the zeroes of v ( t ) = − 9 t 2 + 6 t + 1. Determine the zeroes of f ( x ) = 2 x 2 − 12 x + 23. t = − 6 ± √ ( − 12) 2 − 4(2)(23) 6 2 − 4( − 9)(1) � x = 12 ± 2( − 9) 2(2) = − 6 ±√ 36+36 = 12 ± √ 144 − 184 − 18 √ = − 6 ± 72 4 − 18 = 12 ± √− 40 √ = − 6 ± 6 2 − 18 4 √ = 1 ± 2 3 Since the square root of a negative number is not defined in √ √ the real number system, f ( x ) has no zeroes. The zeroes of v ( t ) are t = 1+ 2 and t = 1 − 2 . 3 3 J. Garvin — Solving Quadratic Equations J. Garvin — Solving Quadratic Equations Slide 13/18 Slide 14/18 f u n c t i o n s f u n c t i o n s The Number/Nature of the Zeroes The Number/Nature of the Zeroes We do not need to use the entire quadratic formula to In the first case, the “plus or minus” in the quadratic determine this. The section inside of the square root is equation ensures that there are two distinct answers. sufficient. In the second case, adding or subtracting the square root of The discriminant of the quadratic formula is b 2 − 4 ac , and zero results in the same answer. can be used to identify the number, and nature, of the zeroes In the last case, there are no real roots because of the of a quadratic function. negative inside of the square root. Using the Discriminant When we are only interested in how many real roots there Given a quadratic function in standard form, are, or whether they are real or complex, we can use the f ( x ) = ax 2 + bx + c , then the function has: discriminant instead of the quadratic formula. • 2 distinct real roots when b 2 − 4 ac > 0 • 1 repeated real root when b 2 − 4 ac = 0 • 2 distinct complex roots when b 2 − 4 ac < 0 J. Garvin — Solving Quadratic Equations J. Garvin — Solving Quadratic Equations Slide 15/18 Slide 16/18 f u n c t i o n s f u n c t i o n s The Number/Nature of the Zeroes Questions? Example State the number, and nature, of the zeroes of g ( x ) = − 5 x 2 + 2 x + 8. The discriminant is 2 2 − 4( − 5)(8) = 164, so g ( x ) has two distinct real roots. Example State the number, and nature, of the zeroes of h ( x ) = 4 x 2 − 12 x + 9. The discriminant is ( − 12) 2 − 4(4)(9) = 0, so h ( x ) has one repeated real root. � 3 Note that h ( x ) = (2 x − 3) 2 , with a vertex at � 2 , 0 . The vertex is the parabola’s only x -intercept. J. Garvin — Solving Quadratic Equations J. Garvin — Solving Quadratic Equations Slide 17/18 Slide 18/18