 # LIAL HORNSBY SCHNEIDER

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LIAL HORNSBY SCHNEIDER
COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER

Completing the Square The Quadratic Formula Solving for a Specified Variable The Discriminant

An equation that can be written in the form where a, b, and c are real numbers with a ≠ 0, is a quadratic equation. The given form is called standard form.

Second-degree Equation
A quadratic equation is a second-degree equation. This is an equation with a squared variable term and no terms of greater degree.

Zero-Factor Property If a and b are complex numbers with ab = 0, then a = 0 or b = 0 or both.

Solve Solution: USING THE ZERO-FACTOR PROPERTY Example 1 Standard form

Solve Solution: USING THE ZERO-FACTOR PROPERTY Example 1
Solve each equation.

Square-Root Property A quadratic equation of the form x2 = k can also be solved by factoring Subtract k. Factor. Zero-factor property. Solve each equation.

Square Root Property If x2 = k, then

Square-Root Property That is, the solution of
Both solutions are real if k > 0, and both are imaginary if k < 0 If k = 0, then this is sometimes called a double solution. If k < 0, we write the solution set as

USING THE SQUARE ROOT PROPERTY Example 2 Solve each quadratic equation. a. Solution: By the square root property, the solution set is

USING THE SQUARE ROOT PROPERTY Example 2 Solve each quadratic equation. b. Solution: Since the solution set of x2 = − 25 is

USING THE SQUARE ROOT PROPERTY Example 2 Solve each quadratic equation. c. Solution: Use a generalization of the square root property. Generalized square root property. Add 4.

Solving A Quadratic Equation By Completing The Square
To solve ax2 + bx + c = 0, by completing the square: Step 1 If a ≠ 1, divide both sides of the equation by a. Step 2 Rewrite the equation so that the constant term is alone on one side of the equality symbol. Step 3 Square half the coefficient of x, and add this square to both sides of the equation. Step 4 Factor the resulting trinomial as a perfect square and combine like terms on the other side. Step 5 Use the square root property to complete the solution.

Solve x2 – 4x –14 = 0 by completing the square.
USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 3 Solve x2 – 4x –14 = 0 by completing the square. Solution Step 1 This step is not necessary since a = 1. Step 2 Add 14 to both sides. Step 3 add 4 to both sides. Step 4 Factor; combine terms.

Solve x2 – 4x –14 = 0 by completing the square.
USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 3 Solve x2 – 4x –14 = 0 by completing the square. Solution Step 4 Factor; combine terms. Step 5 Square root property. Take both roots. Add 2. Simplify the radical. The solution set is

Solve 9x2 – 12x + 9 = 0 by completing the square.
USING THE METHOD OF COMPLETING THE SQUARE a ≠ 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Divide by 9. (Step 1) Add – 1. (Step 2)

Solve 9x2 – 12x + 9 = 0 by completing the square.
USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Factor, combine terms. (Step 4) Square root property

Solve 9x2 – 12x + 9 = 0 by completing the square.
USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Square root property Quotient rule for radicals Add ⅔.

Solve 9x2 – 12x + 9 = 0 by completing the square.
USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Add ⅔. The solution set is

The Quadratic Formula The method of completing the square can be used to solve any quadratic equation. If we start with the general quadratic equation, ax2 + bx + c = 0, a ≠ 0, and complete the square to solve this equation for x in terms of the constants a, b, and c, the result is a general formula for solving any quadratic equation. We assume that a > 0.

Quadratic Formula The solutions of the quadratic equation ax2 + bx + c = 0, where a ≠ 0, are

Caution Notice that the fraction bar in the quadratic formula extends under the – b term in the numerator.

Solve x2 – 4x = – 2 Solution: Here a = 1, b = – 4, c = 2
USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 – 4x = – 2 Solution: Write in standard form. Here a = 1, b = – 4, c = 2 Quadratic formula.

The fraction bar extends under – b.
USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 – 4x = – 2 Solution: Quadratic formula. The fraction bar extends under – b.

The fraction bar extends under – b.
USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 – 4x = – 2 Solution: The fraction bar extends under – b.

Factor first, then divide.
USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 – 4x = – 2 Solution: Factor out 2 in the numerator. Factor first, then divide. Lowest terms. The solution set is

Use parentheses and substitute carefully to avoid errors.
USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Example 6 Solve 2x2 = x – 4. Solution: Write in standard form. Quadratic formula; a = 2, b = – 1, c = 4 Use parentheses and substitute carefully to avoid errors.

Solve 2x2 = x – 4. Solution: The solution set is Example 6
USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Example 6 Solve 2x2 = x – 4. Solution: The solution set is

Cubic Equation The equation x3 + 8 = 0 that follows is called a cubic equation because of the degree 3 term. Some higher-degree equations can be solved using factoring and the quadratic formula.

Solve Solution or or SOLVING A CUBIC EQUATION Example 7
Factor as a sum of cubes. or Zero-factor property or Quadratic formula; a = 1, b = – 2, c = 4

Solve Solution SOLVING A CUBIC EQUATION Example 7 Simplify.
Simplify the radical. Factor out 2 in the numerator.

SOLVING A CUBIC EQUATION
Example 7 Solve Solution Lowest terms

Goal: Isolate d, the specified variable.
SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve for the specified variable. Use  when taking square roots. a. Solution Goal: Isolate d, the specified variable. Multiply by 4. Divide by .

See the Note following this example.
SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. a. Solution Divide by . Square root property See the Note following this example.

Solve the specified variable. Use  when taking square roots.
SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. a. Solution Square root property Rationalize the denominator.

Solve the specified variable. Use  when taking square roots.
SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. a. Solution Rationalize the denominator. Multiply numerators; multiply denominators.

Solve the specified variable. Use  when taking square roots.
SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. a. Solution Multiply numerators; multiply denominators. Simplify.

Solving for a Specified Variable
Note In Example 8, we took both positive and negative square roots. However, if the variable represents a distance or length in an application, we would consider only the positive square root.

Solve the specified variable. Use  when taking square roots.
SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. b. Solution Write in standard form. Now use the quadratic formula to find t.

Solve the specified variable. Use  when taking square roots.
SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. b. Solution a = r, b = – s, and c = – k

Solve the specified variable. Use  when taking square roots.
SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. b. Solution a = r, b = – s, and c = – k Simplify.

The Discriminant The Discriminant The quantity under the radical in the quadratic formula, b2 – 4ac, is called the discriminant. Discriminant

The Discriminant Discriminant Number of Solutions Type of Solution
Positive, perfect square Two Rational Positive, but not a perfect square Irrational Zero One (a double solution) Negative Nonreal complex

Caution The restriction on a, b, and c is important
Caution The restriction on a, b, and c is important. For example, for the equation the discriminant is b2 – 4ac = = 9, which would indicate two rational solutions if the coefficients were integers. By the quadratic formula, however, the two solutions are irrational numbers,

For 5x2 + 2x – 4 = 0, a = 5, b = 2, and c = – 4. The discriminant is
USING THE DISCRIMINANT Example 9 Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. a. Solution For 5x2 + 2x – 4 = 0, a = 5, b = 2, and c = – 4. The discriminant is b2 – 4ac = 22 – 4(5)(– 4) = 84 The discriminant is positive and not a perfect square, so there are two distinct irrational solutions.

First write the equation in standard form as
USING THE DISCRIMINANT Example 9 Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. b. Solution First write the equation in standard form as x2 – 10x + 25 = 0. Thus, a = 1, b = – 10, and c = 25, and b2 – 4ac =(– 10 )2 – 4(1)(25) = 0 There is one distinct rational solution, a “double solution.”

For 2x2 – x + 1 = 0, a = 2, b = –1, and c = 1, so
USING THE DISCRIMINANT Example 9 Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. c. Solution For 2x2 – x + 1 = 0, a = 2, b = –1, and c = 1, so b2 – 4ac = (–1)2 – 4(2)(1) = –7. There are two distinct nonreal complex solutions. (They are complex conjugates.)