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The Humongous Book of Algebra Problems by W. Michael Kelley

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Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
300
Rationalize by multiplying the numerator and denominator by .
The solution to the equation is , , or .
Completing the Square
Make a trinomial into a perfect square
14.12 Given the quadratic expression x
2
– 8x + n, identify the value of n that makes
the trinomial a perfect square and verify your answer.
Divide the coefficient of x by 2.
8 ÷ 2 = –4
Square the quotient.
(–4)
2
= 16
Substituting n = 16 into the expression creates the perfect square x
2
– 8x + 16. To
verify that the quadratic is a perfect square, factor it.
x
2
8x + 16 = (x – 4)(x – 4) = (x – 4)
2
14.13 Given the expression x
2
– 3x + b, identify the value of b that makes the
trinomial a perfect square.
Apply the technique outlined in Problem 14.12: divide the coefficient of x by 2
(that is, multiply it by ) and square the result.
14.14 Solve the equation by completing the square: x
2
– 6x = 0.
Convert the quadratic binomial x
2
– 6x into a perfect square quadratic trinomial
by dividing the coefficient of x by 2 (–6 ÷ 2 = –3) and squaring the result
([–3]
2
= 9).
x
2
– 6x + 9 = 0 + 9
Notice that you cannot add 9 only to the left side of the equation. Any real
number added to one side of the equation must be added to the other side
as well.
x
2
– 6x + 9 = 9
The equation
is allowed to
have up to three
real number solutions
because the highest
power of x in the
original equation
was 3.
x
2
8x + 16
is a perfect
square because it’s
equal to some quantity
multiplied times itself:
(x – 4)(x – 4). It’s the
same logic that makes
25 a perfect square:
5
˙
5 = 25.
This trick only
works if you have
an x
2
-term with a
coefcient of 1
and an x-term.
Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
301
Factor the left side of the equation.
(x – 3)
2
= 9
Solve for x by taking the square root of both sides of the equation. Remember
that this operation requires you to include “±” immediately right of the equal
sign.
Solve both equations that result.
The solution to the equation is x = 0 or x = 6.
14.15 Solve by completing the square: x
2
+ 2x – 7 = 0.
Move the constant to the right side of the equation by adding 7 to both sides.
x
2
+ 2x = 7
Divide the coefficient of x by 2 and square the result: (2 ÷ 2)
2
= 1
2
= 1. Add 1 to
both sides of the equation.
Factor the perfect square.
(x + 1)
2
= 8
Solve for x by taking the square root of both sides of the equation.
Solve for x by subtracting 1 from both sides of the equation.
The solution to the equation is or .
Heres
a tip: When
you factor
the trinomial,
you’ll always get
(x + something)
2
.
That “something” is
half of the original
x-coefcient, the
number you
squared to get 9
a few steps
ago.
The equation
x – 3 = ±3
represents two
different equations:
x – 3 = +3 or
x – 3 = –3.
This leaves
an x
2
-term
and an x-term
on the left side of
the equation, so you
can come up with a
constant to make
a perfect square
like in Problems
14.1214.14.
Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
302
14.16 Solve by completing the square: x
2
– 5x + 1 = 0.
Move the constant to the right side of the equation by subtracting 1 from both
sides.
x
2
– 5x = –1
Square half of the x-coefficient: ; add the result to both
sides of the equation.
Factor the left side of the equation.
Solve the equation for x.
The solution to the equation is or .
Dont forget
the factoring
trick: this number
is always half of
the original x-
coefcient.
You can
add these
fractions because
they have the same
denominator.

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