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Solving Quadratic Equations
Using 4 Different Methods
GREATEST COMMON FACTOR: In order to solve, find the GCF, then factor it out from each of the terms Ex: 2x^2+4x=0 What is found in common with the a and b of this problem? GCF=2x simply take out the GCF... 2x(x+2) = 0 equal both the 2x and the (x+2) to zero and you have your answers 2x=0 x+2= 0 x = 0; x = -2 DIFFERENCE OF 2 SQUARES: When the sum of a number multiplies its difference (a+b)(a-b) then the product is the difference of their squares (a+b)(a-b)=a^2+b^2 Ex: 1. x^2-49 =0 (x+7)(x-7) = 0 Equal both to zero and you have your answers. x+7=0 x-7=0 x=-7 x=7 2. 0= 4a^2-16 How can we simplify this? If you notice both the 4 and 16 can be squared leaving you with... 0= (2a+4)(2a-4) Then equal both equations to zero, divide and simplify. 2a+4=0 2a-4=0 2a=-4 2a=4 a=-4/2 a=4/2 You have your answers. a=-2 a=2 TRINOMIALS: X^2 +bx = c Step 1: Set up a product of two parentheses where each will hold 2 terms (____)(____) Step 2: Find the factors that go in the first positions. To get x^2 an x must go in each of the parentheses (x___)(x___) Step 3: Find the factors that go in the last positions. The two factors will, if multiplied, equal c and, if added, equal b. Ex: x^2+2x-15=0 What 2 numbers can we use that will get us 15 if we multiply and 2 if we add? If you guessed -3 and 5, you would be correct. so (x-3)(x+5) set both equal to zero x-3=0 x+5=0 x=3; x=-5 What happens if you cannot factor so easily? 2x^2+10x+12 As you know we cannot simply factor without a bit of tweaking, so what should we do first? Well, there are two ways as to going about this problem. You may: 1. "Bust the b" 2. Divide out 'a' from the rest of the problem BUST THE B First: multiply your a and c in this case: 2 x 12 =24 Then find factors for both 2 and 12 that equal the b or 10: +6 and +4 your factor will look like this: (2x+4)(2x+6) Divide out the 2, equal each to zero and you're left with x = -2;-3 DIVIDING OUT A 2x^2+10x+12/2 After dividing out the 'a' you should be left with: x^2+5x+6 = 0 (x+2)(x+3) equal both to zero and you get x = -2;-3 Both work just the same. It all depends on the solver's comfort level.
Factor it!
Square root problems generally look like this a^2-c = 0 In order to solve you must first move the c to the other side a^2 = c Then, simply take the square root of both sides. By doing so, the square root and the squared cancel out and you are left with a. a = sqrt(c) Reminder: You cannot square root a negative number. If you happen to come across a problem like this then there is no solution. Ex 1: x^2-25 = 0 Move the 25 to the other side and take the square root of both sides sqrt(x^2) = sqrt(25) The x^2 and the square root cancel until you are left with just x. The square root of 25 is -5 and +5. x = 5; x = -5 Ex 2: x^2+81 = 0 sqrt (x^2) = sqrt(- 81)... CANNOT HAPPEN! No Solution
Square Root
Start out with your basic trinomial: ax^2+bx+c = 0 Step 1: Move c to the right of the equal sign ax^2+bx = c Step 2: take 1/2 of b^2 then add to each side ax^2+bx+[(1/2)(b)]^2 = c + [(1/2)(b)]^2 Step 3: Factor on the left of the equal sign and simplify to the right Step 4: Square root each side and solve. Example: x^2+8x+15 = 0 x^2+8x = -15 x^2+8x+[(1/2) (8)]^2 = 15 +[(1/2)(8)]^2 x^2+8x+16 = -15 +16 (x+4)^2 = 1 sqrt[(x+4)^2] = sqrt(1) x+4 = -1;+1 x = -4 -1 x = -4 +1 x = -5 x = -3
Complete The Square
Start out with a basic trinomial: ax^2+bx+c = 0 Use the formula: x = - b * +;-sqrt(b^2 - 4ac)/2a Example: y = x^2 + 8x + 15 -8+;-sqrt(8^2 - 4(1)(15))/2(1) -8 +;- sqrt(64 - 60)/2 -8 +;- sqrt(4)/2 -8 +;- +2;-2/ 2 -8+2= -6/2 -8-2 = -10 x = -6/2 = -3 x = -10/2 = -5
Remember the Song! la! la! la!
Quadratic Formula
