Pre-statistics using the quadratic
formula to solve quadratic equations
Lesson objective: solve quadratic equations
by using the quadratic formula. Lesson
objective: the quadratic formula. The
solutions of the quadratic equation, ax
square plus bx plus c = 0 are given
by the quadratic formula: x =
negative b plus or minus the square root
of (b squared minus 4ac) all
divided by 2a. So the values of a b and c
are the coefficients for the quadratic
equation in standard form. A common
mistake is to leave out the negative b
when we are dividing by 2a. The 2a is
dividing the negative b plus or minus
the square root of b squared minus 4ac.
Let's do an example, example 1. Solve x
squared minus 6x plus 8 equals 0.
Solution: in this case a is equal to 1, b
is equal to negative 6 and c is equal to
8, here we have the quadratic formula now
all we have to do is substitute these
values in this formula. So b, negative 6
goes here and here, 1 goes here it goes
here, and 1 goes here. Now we follow the
order of operations, we have to compute
this first, therefore we have to compute
negative 6 squared first. Negative 6
squared is 36 minus 4 times 1, which is 4,
times 8 which is 32. A negative negative
6 becomes a positive 6,
and 2 times 1 is 2. Next we want to
subtract and we get 6 plus or minus the
square root of 4 all divided by 2.
Square root of 4 is 2 so we have 6 plus
or minus 2 divided by 2 so this is 2.
Solutions 6 minus 2 divided by 2 and 6
plus 2 divided by 2 6 minus 2 is 4,
divided by 2 is 2 and 6 plus 2 is 8,
divided by 2 is 4. So the solution to
this quadratic equation is x equals 2
and x equals 4. To check it we start with
the original equation for x equals 2
everywhere we see an x we substitute 2.
The left hand side should equal 0
because the right hand side is 0, so we
now have 2 squared minus 6 times 2 plus
8 if 2 is a solution this expression
here will equal 0 we have to square
first 2 squared is 4 6 times 2 is 12
plus 8. Next we subtract we get negative
8 plus 8 and that equals 0 it checks for
x equals 4, the process is essentially
the same, we substitute 4 and for x that
gives us 16 minus 24 plus 8. 16 minus 24
is negative 8, negative 8 plus 8 is 0. It
checks so we have verified the solution
algebraically.
To verify using the graphing calculator
we type in the quadratic equation in y1,
we press "graph" it appears that we have
an x-intercept at 2 and 1 at 4. To check
it with the table, we hit "second graph" we
can see when x is 2, y is 0, when x is 4, y
is 0, so we have verified using the
graphing calculator. There is an
additional way we can verify, that's
using the value feature if you press "2nd"
and "trace" which is "calculate" number one
is value press ENTER you'll see this
screen is wanting to know what value of
x do you want type in a 2 hit enter and
the calculator gives the output for the
y which in this case is 0 and when we do
4 we get the same value of 0. Let's do
another example. Example 2, solve 2x
squared equals 10x minus 3. Solution: our
first step is to take this quadratic
equation and put it in standard form: ax
squared plus bx plus c equals 0, so this
is not equal to 0, this is not in
standard form. So our first step is to
subtract 10x to both sides of the equal
sign and finally add 3 to both sides and
we now have 2x squared minus 10x plus 3
equals 0. So the value of a is 2, the
value of b is negative 10, and the value
of c is 3. Substituting these values in
our formula, negative 10 will go in for b
here, and here a is 2,
three is c, and the denominator is two
times a which will be 2 times 2. Again
working inside the square root negative
10 squared is 100 this will always be a
positive number make sure you include
parenthesis negative 10 times negative
10 is a positive 100 4 times 2 is 8
times 3 is 24. A negative negative 10
becomes a positive 10 and 2 times 2 is 4
so we have x equals 10 plus or minus the
square root of 76 divided by 4 square
root of 76 is not a perfect square, it's
an irrational number, so this is our
exact answer and we can approximate
square root of 76 if we need to to
approximate, we first split the plus and
minus up so the first one is 10 minus
the square root of 76 divided before
using your calculator the square root of
76 is approximately equal to eight point
seven two I take ten minus eight point
seven two I get one point two eight and
finally the last step I divide by four
and we get the approximation of 0.32. The
other half is 10 plus the square root
76 divided by four that becomes ten plus
eight point seven two divided by four
which is eighteen point seven two
divided by four which rounds to four
point six eight so these are the two
approximations.
To check the solutions algebraically I
go to the original equation and
everywhere I see an x I'm going to
substitute our first solution 0.32 the
left-hand side should equal the
right-hand side the right hand side we
have zero,
it's the left-hand side should equal
zero we square first x 2 multiplied by
10 and then combined the three terms
and we get point zero zero four eight
and that is approximately equal to zero
it checks. To check the other solution we
do the same thing, this time we
substitute for point six eight in for x,
following the order of operation we
again get point zero zero four eight and
that's approximately equal to zero it
checks using the graphing calculator we
go to y1 put in our equation hit "graph"
and we can see here are our X intercepts.
Using the table command we can see a
sign change in between 0 & 1 and between
4 & 5, using the value command which is
"2nd" and "trace" then number 1 we can type
in our approximation 0.32 hit enter and
we see the output as point 0 0 for 8 for
the second solution we do the same thing
and again we get an output of point 0 0
4 8
if we use our exact answer which is 10
minus the square root of 76 all divided
by 4 and don't forget your parentheses
here 4 is dividing this whole expression
and hit enter we get a value of 0
for the second solution 10 plus the
square root of 76 all divided by 4 we
get a value that should say zero but
this is negative 1 times 10 to the
negative 12 power which is very very
close to zero so this checks. Recall when
modeling a quadratic situation,
identifying the important parts of a
parabola the x-intercepts the
y-intercept and the vertex can be found
algebraically graphically and
numerically. Using the quadratic formula
is the algebraic way to solve equations
and defined x-intercepts of a parabola.
Thanks for watching!
