With Professor Owl Created by Robbie Smith

Quadratic Term: ax² Linear Term: bx Constant Term: c In order to have a solution, the line or parabola must touch the x-axis once or twice. If it doesn’t touch at all, there is no solution. You must also find the vertex and axis of symmetry

 To find the points, you can make a table! XY Ex. x²+6x+14=y Then graph the results. This graph shows that there is no solution. Vertex: (-3,5) Axis or Sym: -3 Solution: None

 Remember: Most equations can be factored, but not all equations can be factored.  Here are some examples of factoring. Ex. (x+5)(3x-2)=0 X+5=0 3x-2= X=-53x=2 3 x=2 3 Unlike the last example which was already factored, you must factor this one. Ex. 16x² -9=0 (4x+3)(4x-3)=0 4x+3=0 4x-3= x=-3 4x= X=-3 x=3 4 4

 More Examples Set the equation equal to zero Ex. x²+16=8x -8x -8x x²-8x+16=0 (x-4)(x-4)=0 x-4= x=4 Write in Proper Form Ex. 8x² -2=-2x +2x +2x 8x² -2=0 x²-16=0 (8x+4)(8x-4) 8 (x+4)(8x-4) x=-4 x=1 2 Ex. 3x²+6x=0 3x(x+2)=0 x= 0 x= -2

 Examples of Perfect Squares x²+4x+4 (x+2)(x+2)=(x+2)² x²-8+16 (x-4)(x-4)=(x-4)² x²+14x+49 (x+7)(x+7)=(x+7)² To find the constant, do the following: Divide Linear by 2 Then square it. x²+4x+? 4/2=2 2²=4 x²+4x+4

 Examples x²+4-10= x²+4x=10 4/2=2 2²=4 x²+4x+4=10+4 (x+2)²=14 √(x+2)²=√14 x+2=√14 x=-2+√14 x=-2-√14 3x²+6x-9= x²+2x-3=0 x²+2x+1=3+1 √(x+1)=√4 x+1=2 x+1=-2 x=1 x=-3

 When you get an equation, it looks like this: ax²+bx+c When using the quadratic formula, use this formula: x=-b±√b²-4ac 2a Let’s see an example! The Discriminant: b²-4ac (Very Important) It tells you the numbers, root, and solutions. Sweet!

Discriminant: Negative-2 Imaginary Solutions Zero- 1 Real Solution Positive-perfect Square- 2 Reals Rational Positive-Non-perfect square- 2 Reals Irrational You must find the discriminant ! 3x²+6x-9=0 6²-4(3)(-9) = : Two Reals Rational x=-6±12 2(3) x= x= x=1x=-3