Solving Quadratic Equations
Solving quadratic equations means to find when the branches
crosses the x axis. Completing the square finds the max/min point (vertex) which
can be helpful for some types of questions, but other types of question
ask about points that cross the axis.
In order to solve the problem:
1. there are several ways to start this process. In grade ten
you learned two:
a. Factoring
i. Using simple or complex factoring techniques find the points
and
solve for 0 = (x + a)(x + b)
b. Quadratic Equation
i. The plug and play solving method
c. The new method is to us completing the square to solve the
equation for the roots
i. Complete the square so that it is in the form 0 = a (x + h)^{
2} + k
ii. arrange the equation so that the square is on one side and
the constant is on the other.
iii. Solve for the variable using exponent laws (since it is
always squared, both sides are square rooted)
iv. Isolate for the variable
Possible outcomes for roots of a Quadratic equation:
1. 2 real roots
a. notice the values of the coeffiecients
i. if the vertex is above the x axis AND a is negative,
then 2 real
roots.
ii. If the vertex is below the x axis and the value of a
is positive, then
there are two real roots
2. 1 real root
a. notice there is no vertical displacement. If the vertex is on
the x axis there is only 1 real root
3. no real roots (therefore two complex roots)
a. notice the location of the vertex
i. the vertex is above the x axis and the parabola points up,
there are no real roots (the graph never crosses the x axis)
ii. the vertex is below the x axis and the parabola points down,
there are no real roots. (the graph never crosses the x axis)
Example 1
kx  8 = 2x^{ 2}
What values does k have for two distinct real roots, one real
root or no real roots.
Two real roots
2x^{ 2 } +
kx  8 > 0
k^{ 2}  4(2)(8) > 0
k^{ 2} > 64
k > 8; k < 8
one root
2x^{ 2 } +
kx  8 = 0
k^{ 2}  4(2)(8) = 0
k^{ 2} = 64
k = 8; k = 8
no roots
2x^{ 2 } +
kx  8 < 0
k^{ 2}  4(2)(8) < 0
k^{ 2} < 64
k < 8; k > 8
