Quadratic Equations
A
quadratic equation is an equation with unknown variable to the
second power.
The
standard form of a quadratic equation is
Where
a, b, and c are constants and x
is the unknown variable.
If an
equation is not in standard form, we must manipulate it until it
is.
Whenever
we set y = 0 in any function, we are finding the x–intercept(s)
for that function.
So when we
solve a quadratic in the form we are really finding the x–intercepts.
These x–intercepts
are also called roots.
Another
term for roots or x-intercepts is zeros.
So, when
we solve a quadratic in the standard form we are finding the
roots of the function.
Roots
and zeros are very important in electronics and control
systems.
Solving Quadratic
Equations
There are a variety of
techniques for solving quadratic equations including:
Factorization;
Using the quadratic formula; Completing the square; Using a
graph.
We will discuss the first
two techniques only.
Solving by Factorization
A number is said to be
factorized when it is written as a product.
For example 24 can be
factorized into 8 ´
3. Algebraic expressions
can also be factorized, as given
Example:
Factorize z2
– 5z.
Solution
Solving Using a Formula
Not
all quadratics can be factorized. Accordingly, we will present the
quadratic formula for solving the above equation. The quadratic
formula is
Consider
the following quadratic equation
Values
of constants are: a = 1, b = 5, and c =
6.
These
values can be substituted into the quadratic formula to give
There are
two values of x:
the first
one corresponds to the + sign and the second one correspond to
the – sign. The answer is
Both
values satisfy the original equation.
Example:
The
bending moment of a beam M, is given by the equation
where
x is the distance (m) along a beam from one end. Find
the value of x for which M = 0.
Solution: We have
We
use to determine x.
Values
of constants are: a = 7, b = 4, and c =
-3, which can be substituted into the quadratic equation to
give
There
are two values of x:
the
first one corresponds to the + sign and the second one
correspond to the – sign.
The
answer is
Since
we cannot have a distance of -1.00 on the beam, the bending
moment M = 0 is at x = 0.429 m.
