Solving Quadratic Equations Pre-Calc Lesson 1.6 Solving Quadratic Equations Quadratic Equation - Any equation that can be written in ax2 + bx + c = 0 form. **Three methods for solving quadratic equations: 1. Factoring 2. Completing the square!  3. Quadratic formula!  Example 1: Solve by factoring (3x – 2)(x + 4) = - 11 Since this is not = 0, we must first expand by using ‘foil’ 3x2 + 12x – 2x – 8 = - 11 + 11 + 11 3x2 + 10x + 3 = 0 Factor: (3x + 1)(x + 3) = 0 3x + 1 = 0 and x + 3 = 0 x = -⅓ & x = - 3

Example 2: Solve by Completing the Square! 2x2 – 12x – 7 = 0 ??? 1st divide all by 2 so coefficient of x2 is 1 x2 – 6x – 7/2 = 0 Now isolate the ‘x’ terms  add 7/2 to both sides x2 - 6x = 7/2 Now take ½ of ‘b’, then ‘square it and add that result to both sides. -6/2 = (-3)2 = 9 x2 – 6x + 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

Example 3: Solve by using the Quadratic Formula!!! 2x2 + 7 = 4x 1st get in ax2 + bx + c = 0 order 2x2 – 4x + 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 x = 4 + 2i√10 4 now factor a ‘2’ from all three terms and get: x = 2 + i√10 2

Discriminant. In the Quadratic Formula we have the expression under the radical b2 – 4ac 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: b2 – 4ac < 0 - there will exist 2 complex conjugate roots b2 – 4ac = 0 - There will only exist 1 real root – called a double root 3. b2 – 4ac > 0 - There will exist ‘2’ ‘distinct’ real roots

Some Helpful Hints: Quadratic Formula!!  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: If a,b, & c are integers, and if b2 – 4ac is a perfect square, then factor the sucker! If the equation has the form: x2 + (even #)x + c = 0, then solve by completing the square!! If ‘niether’ of those two cases exist, then use the Quadratic Formula!! 

Two special circumstance to look for: (Possibly losing a root! ) Take the equation: 4x(x – 1) = 3(x – 1)2 One approach might be: Divide both sides by (x – 1)  4x = 3(x – 1) 4x = 3x – 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 – 4x x – 2 x + 2 x2 – 4 1st – 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)[ 8 – 4x ] x – 2 x + 2 (x – 2)(x + 2) (x + 2)(x + 2) + (x – 2)(x – 2) = 8 – 4x x2 + 4x + 4 + x2 – 4x + 4 = 8 – 4x  4x’s cancel 2x2 + 8 = 8 – 4x  8’s xancel 2x2 + 4x = 0  factor out a ‘2x’

2x(x + 2) = 0  soooooo 2x = 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!