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: √ 2 x2−4 2 x + 5 = 9 x + 5x = 2x − 2 x+1 = 0 x + 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:

3x 6 Replace x−1 = x−1 , x 6= 1

3x 6 by x−1 ∗ (x − 1) = x−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 x − 3 = 0

1 Example:

8x − (3x + 2) = 3x − 10 8x − 3x − 2 = 3x − 10 5x − 2 = 3x − 10 2x = −8 x = −4

Example:

2 1 1 3 p = 2 p + 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

−15 x = 7 Example:

5 4 9 y+2 − y−6 = y+4 5(y − 6)(y + 4) + 4(y + 2)(y + 4) = 9(y + 2)(y − 6) 5(y2 − 2y − 24) + 4(y2 + 6y + 8) = 9(y2 − 4y − 12) 5y2 − 10y − 120 + 4y2 + 24y + 32 = 9y2 − 36y − 108 14y − 88 = −36y − 108 50y = −20

2 y = 5

2 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}

3 Example 5: Solve the equation 3x2 − 7x − 6 = 0. ⇐⇒ (3x + 2)(x − 3) = 0 ⇐⇒ 3x + 2 = 0 or x − 3 = 0

2 ⇐⇒ x = − 3 or x = 3. 2 So the solution in set notation is x ∈ {− 3 , 3}

Recall: For a quadratic function in standard form, the quadratic formula is √ −b± b2−4ac x = 2a

Example 6: Solve 3x2 − 5x + 1 = 0.

⇐⇒ 3x2 + (−5)x + 1 = 0 ⇒ a = 3, b = −5, c = 1

√ √ √ 5± 52−4(3)(1) 5± 25−12 5± 13 So x = 2(3) = 6 = 6 . √ √ n 5− 13 5+ 13 o Thus x ∈ 6 , 6

Note: If b2 − 4ac < 0, then the quadratic equation has no (real) solutions.