Example 1 Solve each of the
following equations by factoring.
(a) x2 − x =12
(b) 4m2
−1 = 0
Solution
Now,
as noted earlier, we won’t be putting any detail into the factoring process, so
make sure that you can do the factoring here.
(a) x2 − x =12
First, get
everything on side of the equation and then factor.
x2 – x – 12 = 0
(x – 4)(x + 3) = 0
Now
at this point we’ve got a product of two terms that is equal to zero. This
means that at least one of the following must be true.
x – 4 = 0 OR x + 3 = 0
x = 4
OR x
= – 3
Note
that each of these is a linear equation that is easy enough to solve. What this
tell us is that we have two solutions to the equation, x = 4 and x = − 3. As with
linear equations we can always
check our solutions by plugging the solution back into the equation. We will
check x = − 3 and leave the
other to you to check.
So, this was in fact a solution.
(b) 4m2
– 1 = 0
As
always let’s first factor the equation.
4m2 – 1 = 0
(2m – 1)(2m + 1) = 0
Now
apply the zero factor property. The zero factor property tells us that,
2m – 1 = 0 OR 2m + 1 = 0
2m = 1 OR 2m = – 1
m = ½ OR m = – ½
Again,
we will typically solve these in our head, but we needed to do at least one in
complete detail.
So we have two solutions to the equation.
m = ½ AND m = – ½ .
Example 2 Solve each of
the following equations.
Solution
Okay,
just like with the linear equations the first thing that we’re going to need to
do here is to clear the denominators out by multiplying by the LCD. Recall that
we will also need to note value(s) of x that will give division by zero
so that we can make sure that these aren’t included in the solution.
The
LCD for this problem is (x + 1)(2x − 4) and we will
need to avoid x = −1 and x = 2 to make sure
we don’t get division by zero. Here is the work for this equation.
2x – 4 =
(x + 1)(2x – 4) – 5(x + 1)
2x – 4 =
2x2 – 2x – 4 – 5x – 5
0
= 2x2 – 9x – 5
0 = (2x + 1)(x – 5)
So,
it looks like the two solutions to this equation are,
x = – ½ and x = 5
Notice
as well that neither of these are the values of x that we needed to
avoid and so both are solutions.
In this case
the LCD is x − 1 and we will
need to avoid x = 1 so we don’t
get division by zero.
Here is the work for this problem.
(x
– 1)(x + 3) + 3 = 4 – x
x2 + 2x
– 3 + 3 = 4 – x
x2 + 3x
– 4 = 4
(x – 1)(x + 4) = 0
So,
the quadratic that we factored and solved has two solutions, x =1 and x = − 4. However, when
we found the LCD we also saw that we needed to avoid x =1 so we didn’t
get division by zero. Therefore, this equation has a single solution,
x = − 4.
Example 3 Complete
the square on each of the following.
(a) x2 − 16x
(b) y2 + 7y
Solution
(a) x2 −16x
Here’s the number that we’ll add to the equation.
Notice
that we kept the minus sign here even though it will always drop out after we
square things. The reason for this will be apparent in a second. Let’s now
complete the square.
x2 −16x + 64 = ( x − 8)2
Now, this is a quadratic that hopefully you can factor fairly
quickly. However notice that it will always factor as x plus the blue
number we computed above that is in the parenthesis (in our case that is -8).
This is the reason for leaving the minus sign. It makes sure that we don’t make
any mistakes in the factoring process.
(b) y2 + 7y
Here’s the number we’ll need this time.
It’s a fraction and that will happen fairly often with these so
don’t get excited about it. Also, leave it as a fraction. Don’t convert to a decimal. Now complete
the square.
This one is not so easy to factor. However, if you again recall
that this will ALWAYS factor as y plus the blue number above we don’t
have to worry about the factoring process.
Example 4 Solve each of
the following equations.
Solution
(a) There are two ways to work this
one. We can either leave the fractions in or multiply by the LCD (10 in this
case) and solve that equation. Either way will give the same answer. We will only
do the fractional case here since that is the point of this problem. You should
try the other way to verify that you get the same solution.
In this case
here are the values for the quadratic formula as well as the quadratic formula
work for this equation.
a
= ½
b = 1 c = – 1/10
In
these cases we usually go the extra step of eliminating the square root from
the denominator so let’s also do that,
If you do clear
the fractions out and run through the quadratic formula then you should get exactly
the same result. For the practice you really should try that.
(b) In this case do
not get excited about the decimals. The quadratic formula works in exactly the same
manner. Here are the values and the quadratic formula work for this problem.
a = 0.04 b = −0.23 c = 0.09
Now, to this
will be the one difference between these problems and those with integer or fractional
coefficients. When we have decimal coefficients we usually go ahead and figure
the two individual numbers. So, let’s do that,
Notice that we
did use some rounding on the square root.
Sumber
Sumber
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