1. Solving Equations with the Quadratic Formula By completing the square once for the general equation a x 2 + b x + c = 0, you can develop a formula that gives the solutions of any quadratic equation. The formula for the solutions is called the quadratic formula. THE QUADRATIC FORMULA Let a, b, and c be real numbers such that a  0. The solutions of the quadratic equation a x 2 + b x + c = 0 are: Remember that before you apply the quadratic formula to a quadratic equation, you must write the equation in standard form, a x 2 + b x + c = 0. x = – b ± b 2 – 4ac 2a

2. Solve 2x 2 + x = 5. SOLUTION Write original equation. 2x 2 + x = 5 2x 2 + x – 5 = 0 Write in standard form. Quadratic formula a = 2, b = 1, c = – 5 Simplify. Solving a Quadratic Equation with Two Real Solutions x = –b ± b 2 – 4ac 2a x = –1 ± 1 2 – 4(2)(–5) 2(2) x = –1 ± 41 4

3. Solve 2x 2 + x = 5. SOLUTION The solutions are:  1.35 C HECK Graph y = 2 x 2 + x – 5 and note that the x -intercepts are about 1.35 and about –1.85.  –1.85 Solving a Quadratic Equation with Two Real Solutions and x = –1 – 41 4 –1 + 41 4 x =

4. Solving a Quadratic Equation with One Real Solution Solve x 2 – x = 5x – 9. SOLUTION Write original equation. x 2 – x = 5x – 9 x 2 – 6x + 9 = 0 a = 1, b = –6, c = 9 Quadratic formula Simplify. x = 3 The solution is 3. Simplify. x = 6 ± 0 2 x = 6 ± (– 6) 2 – 4(1)(9) 2(1)

5. Solving a Quadratic Equation with One Real Solution Solve x 2 – x = 5x – 9. C HECK Graph y = x 2 – 6 x + 9 and note that the only x -intercept is 3. Alternatively, substitute 3 for x in the original equation. 6 = 6 3 2 – 3 = 5(3) – 9 ? 9 – 3 = 15 – 9 ?

6. Solving a Quadratic Equation with Two Imaginary Solutions Solve –x 2 + 2x = 2. SOLUTION Write original equation. –x 2 + 2x = 2 –x 2 + 2x – 2 = 0 a = –1, b = 2, c = –2 Quadratic formula Simplify. x = 1 ± i x = –2 ± 2 i –2 Write using the imaginary unit i. The solutions are 1 + i and 1 – i. Simplify. x = –2 ± – 4 –2 x = –2 ± 2 2 – 4(–1)(–2) 2(–1)

7. Solving a Quadratic Equation with Two Imaginary Solutions Solve –x 2 + 2x = 2. SOLUTION C HECK Graph y = –x 2 + 2 x – 2 and note that there are no x -intercepts. So, the original equation has no real solutions. To check the imaginary solutions 1 + i and 1 – i, substitute them into the original equation. The check for 1 + i is shown. 2 = 2 –(1 + i ) 2 + 2(1 + i ) = 2 ? –2i + 2 + 2i = 2 ?

8. In the quadratic formula, the expression b 2 – 4 a c under the radical sign is called the discriminant of the associated equation a x 2 + b x + c = 0. Consider the quadratic equation a x 2 + b x + c = 0. NUMBER AND TYPE OF SOLUTIONS OF A QUADRATIC EQUATION discriminant You can use the discriminant of a quadratic equation to determine the equation’s number and type of solutions. If b 2 – 4 a c > 0, then the equation has two real solutions. Solving Equations with the Quadratic Formula If b 2 – 4 a c = 0, then the equation has one real solution. If b 2 – 4 a c < 0, then the equation has two imaginary solutions. –b ± b 2 – 4ac 2a x =

9. Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. SOLUTION x 2 – 6 x + 10 = 0 Two imaginary: 3 ± i a x + b x + c = 0 EQUATIONDISCRIMINANTSOLUTION(S) b 2 – 4a c (–6) 2 – 4(1)(10) = – 4 Using the Discriminant x = –b ± b 2 – 4a c 2a

10. Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. x 2 – 6 x + 10 = 0 Two imaginary: 3 ± i a x + b x + c = 0 EQUATIONDISCRIMINANTSOLUTION(S) b 2 – 4a c (–6) 2 – 4(1)(10) = – 4 Using the Discriminant x 2 – 6 x + 9 = 0(–6) 2 – 4(1)(9) = 0 One real: 3 x 2 – 6 x + 9 = 0 SOLUTION x = –b ± b 2 – 4a c 2a

11. Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. x 2 – 6 x + 10 = 0 Two imaginary: 3 ± i a x + b x + c = 0 EQUATIONDISCRIMINANTSOLUTION(S) b 2 – 4a c (–6) 2 – 4(1)(10) = – 4 Using the Discriminant x 2 – 6 x + 9 = 0(–6) 2 – 4(1)(9) = 0 One real: 3 x 2 – 6 x + 9 = 0x 2 – 6 x + 8 = 0 (–6) 2 – 4(1)(8) = 4 Two real: 2, 4 SOLUTION x = –b ± b 2 – 4a c 2a

12. In the previous example you may have noticed that the number of real solutions of x 2 – 6 x + c = 0 can be changed just by changing the value of c. y = x 2 – 6 x + 10 y = x 2 – 6 x + 9 y = x 2 – 6 x + 8 Graph is above x -axis (no x -intercept) Graph touches x -axis (one x -intercept) Graph crosses x -axis (two x -intercepts) Using the Discriminant A graph can help you see why this occurs. By changing c, you can move the graph of y = x 2 – 6x + c up or down in the coordinate plane. If the graph is moved too high, it won’t have an x -intercept and the equation x 2 – 6 x + c = 0 won’t have a real-number solution.

13. Using the Quadratic Formula in Real Life Previously you studied the model h = –16t 2 + h 0 for the height of an object that is dropped. For an object that is launched or thrown, an extra term v 0 t must be added to the model to account for the object’s initial vertical velocity v 0. Models h = –16 t 2 + h 0 h = –16 t 2 + v 0 t + h 0 Object is dropped. Object is launched or thrown. Labels h = height t = time in motion h 0 = initial height v 0 = initial vertical velocity (feet) (seconds) (feet) (feet per second)

14. Using the Quadratic Formula in Real Life The initial velocity of a launched object can be positive, negative, or zero. v 0 > 0 v 0 < 0 v 0 = 0 If the object is launched upward, its initial vertical velocity is positive (v 0 > 0). If the object is launched downward, its initial vertical velocity is negative (v 0 < 0). If the object is launched parallel to the ground, its initial vertical velocity is zero (v 0 = 0).

15. Solving a Vertical Motion Problem A baton twirler tosses a baton into the air. The baton leaves the twirler’s hand 6 feet above the ground and has an initial vertical velocity of 45 feet per second. The twirler catches the baton when it falls back to a height of 5 feet. For how long is the baton in the air? SOLUTION h = –16 t 2 + v 0 t + h 0 Write height model. h = 5, v 0 = 45, h 0 =6 Since the baton is thrown (not dropped), use the model h = –16t 2 + v 0 t + h 0 with v 0 = 45 and h 0 = 6. To determine how long the baton is in the air, find the value of t for which h = 5. 5 = –16 t 2 + 45t + 6 0 = –16 t 2 + 45t + 1 a = –16, b = 45, c = 1

16. Solving a Vertical Motion Problem A baton twirler tosses a baton into the air. The baton leaves the twirler’s hand 6 feet above the ground and has an initial vertical velocity of 45 feet per second. The twirler catches the baton when it falls back to a height of 5 feet. For how long is the baton in the air? t  – 0.022 or t  2.8 Quadratic formula a = –16, b = 45, c = 1 Use a calculator. SOLUTION Since the baton is thrown (not dropped), use the model h = –16t 2 + v 0 t + h 0 with v 0 = 45 and h 0 = 6. To determine how long the baton is in the air, find the value of t for which h = 5. Reject the solution – 0.022 since the baton’s time in the air cannot be negative. The baton is in the air for about 2.8 seconds. t = – 45 ± 2089 –32