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Solving Equations
An equation in one variable is a statement in which two expressions, at least one of which
containing the variable, are equal. The expressions are called the sides of the equation. Since
an equation is a statement, it may be either true or false, depending on the value of the
variable.
The values of the variable that result in a true statement are called the solutions or
roots of the equation.
Examples:
x+5=9
x2 −4
x+1
x2 + 5x = 2x − 2
=0
√
x2 + 9 = 5
Sometimes equations can have more than one solution:
x2 −4
x+1
= 0 has x = −2 and x = 2 has solutions.
One method for solving equations is to replace the original equation by a succession of
equivalent equations.
Equivalent Equations
The following procedures result in equivalent equations.
1. Interchange the two sides of the equation:
Replace 3 = x by x = 3
2. Simplify the sides of the equation by combing like terms, eliminating parentheses, and
so on:
Replace (x + 2) + 6 = 2x + (x + 1)
by x + 8 = 3x + 1
3. Add or Subtract the same expression on both sides of the equation:
Replace 3x − 5 = 4
by (3x − 5) + 5 = 4 + 5
4. Multiply or divide both sides of the equation by the same nonzero expression:
Replace
by
3x
x−1
3x
x−1
=
6
x−1 ,
∗ (x − 1) =
6
x−1
x 6= 1
∗ (x − 1)
5. If one side of the equation is 0 and the other side can be factored, then we may set
each factor equal to 0:
Replace x(x − 3) = 0
by x = 0
or
1
x−3=0
Example:
8x − (3x + 2) = 3x − 10
8x − 3x − 2 = 3x − 10
5x − 2 = 3x − 10
2x = −8
x = −4
Example:
2
3p
= 21 p +
1
3
4p = 3p + 2
p=2
Example:
(x + 2)(x − 3) = (x + 3)2
x2 − 3x + 2x − 6 = (x + 3)(x + 3)
x2 − x − 6 = x2 + 6x + 9
x2 − x2 − x − 6x = 9 + 6
−7x = 15
x=
−15
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Example:
5
y+2
−
4
y−6
=
9
y+4
5(y − 6)(y + 4) + 4(y + 2)(y + 4) = 9(y + 2)(y − 6)
5(y 2 − 2y − 24) + 4(y 2 + 6y + 8) = 9(y 2 − 4y − 12)
5y 2 − 10y − 120 + 4y 2 + 24y + 32 = 9y 2 − 36y − 108
14y − 88 = −36y − 108
50y = −20
y=
2
2
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Solving Quadratic Equations
Definition: A quadratic equation is an equation equivalent to one of the form:
ax2 + bx + c = 0,
where a, b, and c are real numbers and a 6= 0. A quadratic equation in this form is said to
be in standard form.
Once a quadratic equation is written in standard form, you can solve for x by:
1. factoring ax2 + bx + c into the product of two linear polynomials and then applying
the zero product property, or
2. by using the square root method, or
3. by using the quadratic formula.
Example 1: Solve the equation x2 − 4 = 0.
⇐⇒ x2 = 4
√
⇐⇒ x = ± 4
⇐⇒ x = 2 or x = −2
So the solution in set notation is x ∈ {−2, 2}
Example 2: Solve the equation x2 − 5x = 0.
⇐⇒ x(x − 5) = 0
⇐⇒ x = 0 or x − 5 = 0
⇐⇒ x = 0 or x = 5.
So the solution in set notation is x ∈ {0, 5}
Example 3: Solve the equation x2 + 5x = −6.
⇐⇒ x2 + 5x + 6 = 0
⇐⇒ (x + 2)(x + 3) = 0
⇐⇒ x + 2 = 0 or x + 3 = 0
⇐⇒ x = −2 or x = −3.
So the solution in set notation is x ∈ {−2, −3}
Example 4: Solve the equation x2 + 7x + 12 = 0.
⇐⇒ (x + 3)(x + 4) = 0
⇐⇒ x + 3 = 0 or x + 4 = 0
⇐⇒ x = −3 or x = −4.
So the solution in set notation is x ∈ {−3, −4}
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Example 5: Solve the equation 3x2 − 7x − 6 = 0.
⇐⇒ (3x + 2)(x − 3) = 0
⇐⇒ 3x + 2 = 0 or x − 3 = 0
⇐⇒ x = − 23 or x = 3.
So the solution in set notation is x ∈ {− 23 , 3}
Recall: For a quadratic function in standard form, the quadratic formula is
x=
√
−b± b2 −4ac
2a
Example 6: Solve 3x2 − 5x + 1 = 0.
⇐⇒ 3x2 + (−5)x + 1 = 0
√
So x =
5±
Thus x ∈
n
52 −4(3)(1)
2(3)
=
⇒ a = 3, b = −5, c = 1
√
5± 25−12
6
=
√
5± 13
.
6
√
√ o
5− 13 5+ 13
,
6
6
Note: If b2 − 4ac < 0, then the quadratic equation has no (real) solutions.
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