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Solving Quadratic Equations II
In the last lesson, quadratics that equaled zero were solved by, first, factoring,
second, setting each factor to zero, and then solving the resulting linear equations. All of the problems had one common thread: each equation had a zero
on one side of the equation. In this lesson, the question to address is how to
deal with an equation that does not equal zero.
The problem ab = 0 implied that either one or both of the factors had to be
zero. What about a problem ab = 1? Does this also have a simple conclusion?
Try some numbers before continuing.
Consider ab = 1. Let a be any number. Then b = a1 , the reciprocal of that a,
and ab = 1 has an infinite number of solutions. Thus, the only way to solve
a quadratic or polynomial is to first set the equation to zero by moving all
non-zero terms to one side of the equation. A few examples clarify.
Example. Solve x2 − 4x = 12.
x2 − 4x = 12
x2 − 4x − 12 = 12 − 12
x2 − 4x − 12 = 0
(FT3)
2
x − 6x + 2x − 12 = 0
x(x − 6) + 2(x − 6) = 0
(x − 6)(x + 2) = 0 which implies
x − 6 = 0 OR x + 2 = 0
−2 = −2
+6 = +6
x = 6
x = −2
Notice that the 12 was moved from the right side of the equation to the left. A
safe, and wise, strategy is to always move the terms in such a way as to keep
the leading coefficient positive 1 A polynomial can be factored when the leading
coefficient is negative, but it only adds unnecessary steps and difficulty.
1 The leading coefficient is the coefficient (or number) with the variable of highest degree,
or exponent. For example, x3 − 14x + 1 = 0 is a degree 3 equation with a leading coefficient
of 1. And, 2x2 + 3x + 4 = 0 is a degree two equation with a leading coefficient of 2.
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Example. Solve 7x = 15 − 2x2 .
7x
+2x + 7x
2x2 + 7x
2
2x + 7x − 15
2x2 + 7x − 15
2
2x + 10x − 3x − 15
2x(x − 5) + 3(x − 5)
(x − 5)(2x + 3)
2
=
=
=
=
=
=
=
=
15 − 2x2
15 − 2x2 + 2x2
15
15 − 15
0
0
0
0
x−5 = 0
+5 = +5
x =5
−30
10
−3
7
OR 2x + 3
−3
2x
x
= 0
= −3
= −3
= − 23
Notice that the −2x2 and the 15 were moved from the right side of the equation
to the left. The motivation for this was to make the leading term of −2x2 to
+2x2 . Remember, a safe strategy is to always move the terms in such a way
as to keep the leading coefficient positive. A polynomial can be factored when
the leading coefficient is negative, but it only adds an unnecessary steps and
difficulty.
Example. Solve x2 = 9. Be careful. The answer might surprise you.
x2 = 9
x2 − 9 = 0
(x + 3)(x − 3) = 0
x+3 = 0
−3 = −3
x = −3
x−3 = 0
+3 = +3
x = +3
This simple problem actually has two answers. The degree of the polynomial
indicates the number of solutions to the equation. A degree one, or linear, equation has one solution, a degree 2 equation has 2 solutions, a degree 3 equation
has 3 solutions, etc.
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Example. Solve 2x2 + 8(x − 1) = 5x + 1.
2x2 + 8(x − 1) = 5x + 1
2x2 + 8x − 8 = 5x + 1
2x2 + 8x − 5x − 8 − 1 = 0
2x2 + 3x − 9 = 0
2x2 + 6x − 3x − 9 = 0
2x(x + 3) − 3(x + 3) = 0
(x + 3)(2x − 3) = 0
x+3 = 0
−3 = −3
x = −3
2x − 3
+3
2x
x
= 0
= +3
= +3
= 23
Practice Problems
1. x2 + 2x = 15
2. x2 − x = 12
3. 6x2 − x = 2
4. 2x2 − 9x = −9
5. x2 + 4(x − 2) = 4
6. x2 − 3(3x − 4) = −8
7. x2 + 7x − 12 = 3x
8. x2 + 2x − 18 = 5x
9. x2 − 10x = −3x
10. (x + 1)(x − 3) = 12
11. x2 − 20 = 5
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Solutions to Practice Problems.
1. x2 + 2x = 15 =⇒ x2 + 2x − 15 = 0 =⇒ (x + 5)(x − 3) = 0 =⇒ x = −5, 3
2. x2 − x = 12 =⇒ x2 − x − 12 = 0 =⇒ (x − 4)(x + 3) = 0 =⇒ x = −3, 4
3. 6x2 − x = 2 =⇒ 6x2 − x − 2 = 0 =⇒ (3x − 2)(2x + 1) = 0 =⇒ x = − 21 , 23
4. 2x2 − 9x = −9 =⇒ 2x2 − 9x + 9 = 0 =⇒ (x − 3)(2x − 3) = 0 =⇒ x = 3, 23
5. x2 + 4(x − 2) = 4 =⇒ x2 + 4x − 12 = 0 =⇒ (x + 6)(x − 2) = 0 =⇒ x = −6, 2
6. x2 − 3(3x − 4) = −8 =⇒ x2 − 9x + 20 = 0 =⇒ (x − 5)(x − 4) = 0 =⇒ x = 5, 4
7. x2 + 7x − 12 = 3x =⇒ x2 + 4x − 12 = 0 =⇒ (x + 6)(x − 2) = 0 =⇒ x = −6, 2
8. x2 + 2x − 18 = 5x =⇒ x2 − 3x − 18 = 0 =⇒ (x − 6)(x + 3) = 0 =⇒ x = 6, −3
9. x2 − 10x = −3x =⇒ x2 − 7x = 0 =⇒ x(x − 7) = 0 =⇒ x = 0, 7
10. (x + 1)(x − 3) = 12 =⇒ x2 − 2x − 15 = 0 =⇒ (x − 5)(x + 3) = 0 =⇒ x = 5, −3
11. x2 − 20 = 5 =⇒ x2 − 25 = 0 =⇒ (x + 5)(x − 5) = 0 =⇒ x = −5, 5
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