Section 10.2 The Quadratic Formula

Objectives Derive the quadratic formula Solve quadratic equations using the quadratic formula Write equivalent equations to make quadratic formula calculations easier Use the quadratic formula to solve application problems

Objective 1: Derive the Quadratic Formula Quadratic Equation Formula: The solutions of ax2 + bx + c = 0 with a ≠ 0, are given by Read as “x equals the opposite of b plus or minus the square root of b squared minus 4ac, all over 2a.”

Objective 2: Solve Quadratic Equations Using the Quadratic Formula To solve a quadratic equation in x using the quadratic formula, we follow these steps. Write the equation in standard form: ax2 + bx + c = 0. Identify a, b, and c. Substitute the values for a, b, and c in the quadratic formula and evaluate the right side to obtain the solutions.

EXAMPLE 1 Solve 2x2 – 5x – 3 = 0 by using the quadratic formula. Strategy We will begin by comparing 2x2 – 5x – 3 = 0 to the standard form ax2 + bx + c = 0. Why To use the quadratic formula, we need to identify the values of a, b, and c.

EXAMPLE 1 Solve 2x2 – 5x – 3 = 0 by using the quadratic formula. Solution We see that a = 2, b = –5 , and c = –3. To find the solutions of the equation, we substitute these values into the quadratic formula and evaluate the right side.

EXAMPLE 1 Solve 2x2 – 5x – 3 = 0 by using the quadratic formula. Solution To find the first solution, we evaluate the expression using the + symbol. To find the second solution, we evaluate the expression using the – symbol. The solutions are 3 and and the solution set is . Check each solution in the original equation.

Objective 3: Write Equivalent Equations to Make Quadratic Formula Calculations Easier When solving a quadratic equation by the quadratic formula, we can often simplify the calculations by solving a simpler, but equivalent equation.

EXAMPLE 4 For each equation below, write an equivalent equation so that the quadratic formula calculations will be simpler. Strategy We will multiply both sides of each equation by a carefully chosen number. Why In each case, the objective is to find an equivalent equation whose values of a, b, and c are easier to work with than those of the given equation.

EXAMPLE 4 For each equation below, write an equivalent equation so that the quadratic formula computations will be simpler. Solution a. It is often easier to solve a quadratic equation using the quadratic formula if a is positive. If we multiply (or divide) both sides of −2x2 + 4x − 1 = 0 by −1, we obtain an equivalent equation with a > 0. Don’t forget to multiply each term by -1.

EXAMPLE 4 For each equation below, write an equivalent equation so that the quadratic formula computations will be simpler. Solution b. For , two coefficients are fractions: . We can multiply both sides of the equation by their least common denominator, 15, to obtain an equivalent equation having coefficients that are integers. On the left side, distribute the multiplication by 15.

EXAMPLE 4 For each equation below, write an equivalent equation so that the quadratic formula computations will be simpler. Solution c. For 20x2 – 60x − 40 = 0, the coefficients 20, −60, and −40 and have a common factor of 20. If we divide both sides of the equation by their GCF, we obtain an equivalent equation having smaller coefficients. The division by 20 is done term-by-term.

EXAMPLE 4 For each equation below, write an equivalent equation so that the quadratic formula computations will be simpler. Solution d. For 0.03x2 – 0.04x – 0.01 = 0, all three coefficients are decimals. We can multiply both sides of the equation by 100 to obtain an equivalent equation having coefficients that are integers. On the left side, distribute the multiplication by 100.

Objective 4: Use the Quadratic Formula to Solve Application Problems A variety of real-world applications can be modeled by quadratic equations. However, such equations are often difficult or even impossible to solve using the factoring method. In those cases, we can use the quadratic formula to solve the equation.

EXAMPLE 5 Shortcuts. Instead of using the hallways, students are wearing a path through a planted quad area to walk 195 feet directly from the classrooms to the cafeteria. If the length of the hallway from the office to the cafeteria is 105 feet longer than the hallway from the office to the classrooms, how much walking are the students saving by taking the shortcut? Analyze The two hallways and the shortcut form a right triangle with a hypotenuse 195 feet long. We will use the Pythagorean theorem to solve this problem. Assign If we let x = the length (in feet) of the hallway from the classrooms to the office, then the length of the hallway from the office to the cafeteria is (x + 105) feet.

EXAMPLE 5 Shortcuts. …. how much walking are the students saving by taking the shortcut? Form Substituting the lengths into the Pythagorean theorem, we have

EXAMPLE 5 Shortcuts. …. how much walking are the students saving by taking the shortcut? Solve To solve x2 + 105x – 13,500 = 0, we will use the quadratic formula with a = 1, b = 105, and c = –13,500. This is the quadratic formula. Substitute for a, b, and c. Evaluate within the radical: 1052-4(1)(-13,500) = 11,025 + 54,000 = 65,025 Multiply in the denominator. Do the division. Since the length of the hallway can’t be negative, discard the solution -180. Add : -105 + 255 = 150. Subtract: -105 – 255 = -360

EXAMPLE 5 Shortcuts. …. how much walking are the students saving by taking the shortcut? State The length of the hallway from the classrooms to the office is 75 feet. The length of the hallway from the office to the cafeteria is 75 + 105 = 180 feet. Instead of using the hallways, a distance of 75 + 180 = 255 feet, the students are taking the 195-foot shortcut to the cafeteria, a savings of (255 – 195), or 60 feet. Check The length of the 180-foot hallway is 105 feet longer than the length of the 75-foot hallway. The sum of the squares of the lengths of the hallways is 752 + 1802 = 38,025. This equals the square of the length of the 195-foot shortcut: 1952 = 38,025. The result checks.