PreCalc Lesson 1 6 Solving Quadratic Equations Quadratic

Pre-Calc Lesson 1. 6 Solving Quadratic Equations Quadratic Equation - Any equation that can be written in ax 2 + bx + c = 0 form. **Three methods for solving quadratic equations: 1. Factoring 2. Completing the square! 3. Quadratic formula! Example 1: Solve by factoring (3 x – 2)(x + 4) = - 11 Since this is not = 0, we must first expand by using ‘foil’ 3 x 2 + 12 x – 8 = - 11 + 11 Factor: + 11 3 x 2 + 10 x + 3 = 0 (3 x + 1)(x + 3) = 0 3 x + 1 = 0 and x + 3 = 0 x = -⅓ & x=-3

Example 2: Solve by Completing 2 x 2 – 12 x – 7 = 0 the Square! ? ? ? 1 st divide all by 2 so coefficient of x 2 is 1 x 2 – 6 x – 7/2 = 0 Now isolate the ‘x’ terms add 7/2 to both sides x 2 - 6 x = 7/2 Now take ½ of ‘b’, then ‘square it and add that result to both sides. -6/2 = (-3)2 = 9 x 2 – 6 x + 9 = 7/2 + 9 (x – 3)2 = 25/2 take the square root of both sides √(x – 3)2 = √ 25/2 x – 3 = + 5/√ 2 Now rationalize this x – 3 = + 5√ 2 2 so x = 3 + 5√ 2 2

Example 3: Solve by using the Quadratic Formula!!! 2 x 2 + 7 = 4 x 1 st get in ax 2 + bx + c = 0 order 2 x 2 – 4 x + 7 = 0 a = 2, b = - 4, and c = 7 Thus: x = - (-4) + √(-4)2 – 4(2)(7) 2(2) x = 4 + √ 16 – 56 4 x = 4 + √-40 4 x = 4 + 2 i√ 10 4 now factor a ‘ 2’ from all three terms and get: x = 2 + i√ 10 2

In the Quadratic Formula we have the expression under the radical b 2 – 4 ac This is called the Discriminant. Is it so appropriately named because it tells us something special (or discriminates) about our roots! (x-intercepts) It can tell us many things about our solutions. (roots, zeros) For Instance, if: 1. b 2 – 4 ac < 0 - there will exist 2 complex conjugate roots 2. b 2 – 4 ac = 0 - There will only exist 1 real root – called a double root 3. b 2 – 4 ac > 0 - There will exist ‘ 2’ ‘distinct’ real roots

Sometimes when sitting by oneself doing one’s Pre-Calc Homework, one may think, “Self, which method should I use to solve this here quadratic equation? ” Some Helpful Hints: 1. If a, b, & c are integers, and if b 2 – 4 ac is a perfect square, then factor the sucker! 2. If the equation has the form: x 2 + (even #)x + c = 0, then solve by completing the square!! 1. If ‘niether’ of those two cases exist, then use the 2. Quadratic Formula!!

Two special circumstance to look for: (Possibly losing a root! ) Take the equation: 4 x(x – 1) = 3(x – 1)2 One approach might be: Divide both sides by (x – 1) 4 x = 3(x – 1) 4 x = 3 x – 3 x = - 3 !! This will only give us one ‘root’ which may not be wrong, but if we check the answer x = 1, which comes from the (x – 1) factor we divided out in the original problem we would see that it makes the equation true also! Therefore never discard a factor with a variable in it!

Case 2: Gaining a root! Check for Extraneous roots! Example: x + 2 + x – 2 = 8 – 4 x x– 2 x+2 x 2 – 4 1 st – find the least common denominator of all the ‘fractions’ = (x – 2)(x + 2) Now multiply each and every term by the LCD! (x – 2)(x + 2) [x + 2 + x – 2] = (x – 2)(x + 2)[ x– 2 à à x+2 (x + 2) + (x – 2) = 8 – 4 x x 2 + 4 x + 4 + x 2 – 4 x + 4 = 8 – 4 x 2 x 2 + 8 = 8 – 4 x 2 x 2 + 4 x = 0 8 – 4 x ] (x – 2)(x + 2) 4 x’s cancel 8’s xancel factor out a ‘ 2 x’

2 x(x + 2) = 0 soooooo 2 x = 0 and x + 2 = 0 x = 0 and x = - 2 butttt if we try to insert and check x = -2 back in the original equation we will get an undefined expression sew ? ? ? x = - 2 is called an extraneous solution and we must discard it sooooo x = 0 is our only solution! Hw Pg ? ? ? We will get in class on Monday!