Notes

quadratic inequalities

for f(x)= x^2 = vertex is always 0,0. always 9,4,1,0,1,4,9
for f(x)= ax^2 = vertex is always 0,0. the y is 9(a),4(a),1(a),0,1(a),4(a) 9(a)
for f(x)= (x-h)^2 = the x of the vertex is equal to h and the y is 0. the y is 9,4,1,0,1,4,9
for f(x)= x^2+k = vertex is always 0,k (no pun intended). the y is 9+k,4+k,1+k,k,1+k,4+k,9+k

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finding the roots of x

the quadratic formula

-b(sqrt b^2 - 4ac) /2
find a,b and c
simply substitute the variables with the values you have found and solve as usual

completing the sqaure

transpose the third term to the right side
find the value of b. divide it by two then square it. add it to both sides
square root both sides
transpose the second term to the right

Example: x^2+5x+6=0

transpose the third term
=x^2+5x=-6
find b. divide it by 2 and square it. add it to both sides
=x^2+5x+(5/2)^2=-6(5/2)^2
simplify
=x^2+5x+(25/4)=-6+(25/4)
=x^2+5x+(25/4)=25-24/4
=x^2+5x+(25/4)=1/4
square root both sides
=x+(5/2)=1/2
transpose the second term to the right
=x=1-5/2
=x=-4/2
=x=-2

the root is -2

factoring binomials by x (x^2+bx=0)
Example: x^2+4x=0

factor by x
=x(x+4)=0
take x+4. transpose 4 to the right side
=x=0 | x+4 = x=-4
the roots are 0 and -4

Another: 2x^2-8x=0
=x(2x-8)=0
=x=0 | 2x-8=0
=x=0 | x-4=0
=x=0 | x=4
the roots are 0 and 4

factoring trinomials (ax^2+bx+c)

Example: x^2+3x+2=0

factor into two binomials
=(x+1)(x+2)
take x+1. transpose 1 to the right side. take x+2. transpose 2 to the right side.
=x=-1 | x=-2
the roots are -1 and -2

dividing the quadratic equation with one of its roots

Example: x^2+3x+2=0

one of the roots is -1
turn it into an equation by transposing 1 to the left side
it becomes x+1
divide x^2+3x+2 by x+1
x+1 times x is x^2+x. x^2+3x+2-x^2-x = 2x+2
x+1 times 2 is 2x+2. 2x+2-2x-2 = 0
take x and 2 then add it. it becomes x+2
transpose 2 to the right side
x=-2

the other root is -2

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the art of trial and error

basically a method for people who can't find the methods above easy. while this takes longer , it takes less effort

Example: x^2+4x+4=0

substitute x by -1

(-1)^2+4(-1)+4=1. wrong value. move on to -2

substitute x by -2

(-2)^2+4(-2)+4=0. correct value.

the root of x is -2. since the quadratic equation is a perfect square trinomial , it only has one root.