Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
296
Solving Quadratics by Factoring
Use techniques from Chapter 12 to solve equations
14.1 Solve the equation: xy = 0.
According to the zero product property, a product can only equal 0 if at least
one of the factors (in this case either x or y) equals 0. Therefore, xy = 0 if and
only if x = 0 or y = 0.
14.2 Solve the equation: x(x – 3) = 0.
The product of x and (x – 3) is equal to 0. According to the zero product
property (explained in Problem 14.1), one (or both) of the factors must equal 0.
x = 0 or x – 3 = 0
Solve x – 3 = 0 by adding 3 to both sides of the equation.
x = 0 or x = 3
The solution to the equation x(x – 3) = 0 is x = 0 or x = 3.
14.3 Solve the equation: (x + 2)(2x – 9) = 0.
According to the zero product property, the product left of the equal sign only
equals 0 if either (x + 2) or (2x – 9) equals 0. Set both factors equal to 0 and
solve the equations.
The solution to the equation (x + 2)(2x – 9) = 0 is x = –2 or .
Note: Problems 14.4–14.5 demonstrate two different ways to solve the equation x
2
= 16.
14.4 Solve the equation using square roots.
Problems 13.32–13.36 demonstrate that squaring both sides of an equation that
contains a square root eliminates the root. Conversely, taking the square root
of both sides of an equation containing a perfect square eliminates the perfect
square.
You can’t
multiply two
numbers and
get zero unless one
(or both) of those
numbers is 0. That’s a
property that is unique
to 0. For example, if
xy = 4, then there’s
no guarantee that
either x = 4 or
y = 4. You could set
x = 2 and y = 2, or
maybe x = 8 and
y = 0.5.
There
are two
possible
solutions to |
the equation.
Use the word
“or” to separate
them because
plugging either
x = 0 OR x = 3 into
the equation produces
a true statement.
Mathematically,
saying “x = 0 AND
x = 3” means x has to
equal both of those
things at the same
time, and that
doesn’t make
sense.
Back in Chapter 13, you only squared
both sides AFTER you isolated the square
root on one side of the equal sign. Similarly,
only square root both sides when the perfect
square is isolated on one side of the equal sign.
Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
297
Notice the “±” sign. Whenever you take the root of both sides of an equation
and the index of that root is even, you must introduce the ± sign on one side of
the equation, usually the side not containing the variable.
The solution to the equation is x = –4 or x = 4.
Note: Problems 14.4–14.5 demonstrate two different ways to solve the equation x
2
= 16.
14.5 Solve the equation by factoring.
To apply the zero product property (the principle used to solve the equations
in Problems 14.1–14.3), one side of the equation must equal 0. Change the right
side of the equation to 0 by subtracting 16 from both sides.
Factor the left side of the equation, a difference of perfect squares.
(x + 4)(x – 4) = 0
Apply the zero product property by setting each factor equal to 0 and solving
the resulting equations.
This solution echoes the solution generated by Problem 14.4: x = –4 or x = 4.
14.6 Solve the equation: x
2
– 11x + 28 = 0.
The right side of the equation equals 0, so factor the quadratic trinomial.
Because the leading coefficient of x
2
– 11x + 28 is 1, use the techniques
described in Problems 12.35–12.41.
Apply the zero product property and solve the equations that result.
The solution to the equation is x = 4 or x = 7.
You have
to include ± or
you’ll exclude valid
answers. 4 is not the
only squared number
that equals 16. –4
works, too.
See Problems
12.26–12.30 to
practice factoring the
difference of perfect
squares.
The rst
step to solving
a quadratic by
factoring is to make
sure one side of the
equation (usually the
right side) equals 0.
That way, after you
factor, you can use the
zero product property
to set the factors
equal to 0.
The coefcient
of x
2
is 1, so to
factor, you’ll need to
nd two numbers that
add up to –11 and
multiply to give you
+28. The numbers
are –4 and –7.
Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
298
14.7 Solve the equation: x
2
= –7x + 78.
To solve a quadratic equation by factoring, only 0 should appear right of the
equal sign. Add 7x to, and subtract 78 from, both sides of the equation.
Factor the quadratic trinomial.
(x + 13)(x – 6) = 0
Apply the zero product property.
The solution to the equation is x = –13 or x = 6.
14.8 Solve the equation: 10x
2
+ 13x – 3 = 0.
Factor the quadratic trinomial by decomposition, as illustrated by Problems
12.42–12.44.
Apply the zero product property to solve the equation.
The solution to the equation is or .
14.9 Solve the equation: 36x
2
+ 12x + 1 = 0.
Factor the quadratic trinomial by decomposition.
The coefcient
of x
2
is 1, so you
need two numbers
with a sum of 7 and
a product of –78. The
product’s negative, so
the signs are different.
The sum is positive, so the
positive number is bigger.
Try 8 – 1, 9 – 2, 10 – 3, …
until you reach
13 – 6 = 7, because
13(–6) = –78.
Use decomposition
because the co-
efcient of x
2
isn’t 1.
You’re looking for two
numbers that have
a sum of +13 and a
product of –30. The
numbers are 15 and
–2.
Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
299
Apply the zero product property.
The solution to the equation is the double root .
14.10 Solve the equation: 12y
3
+ 324 = 0.
The terms 12y
3
and 324 share a greatest common factor: 12. Factor it out of the
sum.
12(y
3
+ 27) = 0
Apply the zero product property.
Discard the left equation, as 12 ≠ 0.
The solution to the equation is y = –3.
14.11 Solve the equation: 6x
3
– 3x
2
– 4x = –2.
Add 2 to both sides of the equation and factor by grouping.
Apply the zero product property to solve the equation.
There’s
only one
equation
(instead of two
separated by the
word “or”) because
both factors are
equal. There’s no
need to solve the
same equation
twice.
It’s called
a “double root”
because you get this
answer TWICE in the
same problem, as the
factor (6x + 1) is
repeated.
It’s okay to
take the root of a
negative number when
the index of the root
is odd: (–3)(–3)(–3) = –27.
Negative roots with an
even index, however,
are imaginary numbers.
To review factoring
by grouping, look at
Problems 12.20–12.25.
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