we make each bracket equal
zero to find the values
We need to find two numbers that
multiply to give c (which in this case
is -12) and add to give b (which is 1)
3 x -4 = -12
3 + -4 = -1
This is a template for out brackets
To solve, it must always equal zero
EXAMPLE 2
solve 3x² + 7x -6 = 0
in this case a (coefficient of x) is not 1
tn = -18
for this, we need two numbers that multiply
to get -18 ( -6 x 3 ) , and add to get 7
9 and -2
These are our two numbers, 9 x -2 = -18, and 9 + -2 = 7
3x² + 9x -2x -6 = 0
we replace the b value with our two numbers
,3x² + 9x, ,-2x -6,
3x(x + 3)
-2(x + 3)
(3x -2) (x + 3)
x = -3 x = 2/3 / 6.67
you then reverse the brackets and thats your two answers
The two on each outside and put them in
a bracket, and the others in the bracket.
these are out two factorised brackets
we then use the same numbers in the brackets with
the seconds two parts ( x+ 3), and then chose what
these are multiplied by and put it on the outside
we then find a common denominator between
the first two parts, in this case, 3x. we then put
the two numbers inside the brackets
we then separate it into two parts, the
first two bits, and the last two bits
Quadratic formula
-b ± √ b² - 4ac
x=____________________
2a
example:
solve 3x² + 7x = 1
3x² + 7x -1 = 0
x = -7 ±√ 7²-(4 x 3 x -1)
_______________
2 x 3
= -7 ± √ 49 + 12
___________
6
= -7+√61
_______
6
= -7-√61
_______
6
you now have two equations
to put into the calculator
x = 0.1350
x = -2.468
these are the two solutions
4 x 3 x -1 = -12, 7² - -12 is 49 + 12
then we sub in the values into
the formula, a = 3, b = 7 and c = -1
first put the equation into
the standard format
completing the square
convert your equation into the
standard format, ax² + bx + c = 0
write out the initial bracket as (x + b/2)²
multiply out the brackets and
compare it to the original equation
than add or subtract a number to make
it identical to the original equation, add
this value on after the (x + b/2)²
-HOWEVER- if a ≠ 1, before all of these steps, once your
equation is in the standard format divide it by a, and
have a on the outside of the rest of the equation in a
bracket to make the coefficient of x, 1. then continue
with the rest of the steps
write 2x² + 5x + 9 in the form a( x + m)² + n
2x² + 5x + 9
2 (x² + 5/2 x) + 9
2 (x + 5/4)² = 2x² + 5x + 25/8
9 = 72/8
72/8 - 25/8 = 47/8
2 (x + 5/4)² + 47/8
putting that on the end of the initial bracket
finding out the value to add/subtract
to complete the square
converting c to the same
fraction ( to make it easier )
this is the initial bracket,
multiplied out
take out a factor of two
standard format
Example:
Quadratic graphs
quadratic functions are the form y = anything with x²
they all have a symmetrical bucket shape
to complete a table of values from an
equation, you just sub in the x values, and
get the y value to get a point on a graph
when sketching a graph, always
label the important points on it
the solution to a quadratic are
the x intercepts on a graph