When mathematicians use a procedure
repeatedly they often try to find a
formula for the procedure. The quadratic
formula condenses into one calculation
the many steps use to solve a quadratic
equation by completing the square. Here
we have some examples of equations with
second degree terms. We can use the
quadratic formula to solve quadratic
equations. If the quadratic equation is
in the form of ax squared plus bx plus c
equals 0 in other words standard form.
Then we can use the quadratic formula
shown here.
Where a is the coefficient of the
squared term it's underneath the radical
as well as in the denominator, b is the
coefficient from the linear term which
is used in two places in the quadratic
formula. Then c the constant term
which is here shown underneath the
radical as well. Let's take a look at
using the quadratic formula to solve
this quadratic equation. This equation is
in standard form which allows us to
determine the value of a which is 4, b
which is 5 and c which is a negative 6.
Replacing these values for the variables
in the quadratic formula. We now have the
opposite of b negative b plus or minus b
which is 5 squared minus 4 times a which
is 4 times c which is negative 6 all
over 2a and a is 4. Next step then is to
simplify this quadratic equation.
We'll focus our attention on the
quantity underneath the radical. We want
to simplify this expression
into a single value for the radicand
following order of operations will
square the 5 and while we're at it
multiplying a negative 4 times 4
times a negative 6 results in positive
96,adding those two values together
results in the square root of 121.
Continuing to simplify this expression
square root of 121 since it's a perfect
square results in X equal to negative 5
plus 11 or negative 5 minus 11 each over
8 simplifying this combined solution
into the two possible solutions is the
following. Solving then for the variable
we can add the numerator giving us 6
over 8 and reducing results in X equal 3,
4's. In the other solution negative 5
minus 11 results in a simplified or
substituted value of negative 16 over
the denominator of 8 which reduces to X
equal negative 2. Giving us the two
solutions for the quadratic equation.
