Solving Equations with the Quadratic Formula

Transcription

1 0 Solving Equations with the Quadratic Formula In this chapter, you will have the opportunity to practice solving equations using the quadratic formula. In Chapter 17, you practiced using factoring to solve quadratic equations, but factoring is useful only for those equations that can easily be factored. The quadratic formula will allow you to find solutions for any quadratic equation that can be put in the form ax + bx + c = 0, where a, b, and c are numbers. The quadratic formula tells you that for any equation in the form ax + bx + c = 0, the solution will be x = b ± b 4ac. a The solutions found using the quadratic formula are also called the roots of the equation. Some solutions will be in the form of whole numbers or fractions. Some will be in the form of a radical. Some will be undefined, as when the radicand is equal to a negative number. The ± in the quadratic equation tells you that there will be two solutions, one when you add the radical and one when you subtract it.

2 Tips for Solving Equations with the Quadratic Formula Transform the equation into the form ax + bx + c = 0. Use the values for a, b, and c in the quadratic equation to determine the solution for the original equation. For solutions that contain a radical, be sure to simplify the radical as you practiced in Chapter 1. When you are asked to find the solution to the nearest hundredth, you can use a calculator to find the value of the radical. Solve the following equations using the quadratic formula. Reduce answers to their simplest form or to the simplest radical form. Use of a calculator is recommended x + x = x 7x 30 = x + 13x = x + 9x + 1 = x + 17x = x = 1x x + 5x = x = x = 1x x + 11x 7 = x + 5x + 0 = x = 5x 4. x + x = x = 0x x = (x 1) 491. x + x + 11 = x + 1x 6 = x = 4(3x + 1) x x 3 = x 1x + 1 = Algebra Questions 6

3 Find the solution to the following equations to the nearest hundredth r 4r 7 = m + 1m = y = 16y s + 1s 1 = c 11c + = k 3k + = 0 63

4 Answers Numerical expressions in parentheses like this [ ] are operations performed on only part of the original expression. The operations performed within these symbols are intended to show how to evaluate the various terms that make up the entire expression. Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression. Once a single number appears within these parentheses, the parentheses are no longer needed and need not be used the next time the entire expression is written. When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within are to be multiplied. Sometimes parentheses appear within other parentheses in numerical or algebraic expressions. Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first and work outward. The solutions are underlined The equation is in the proper form. First, list the values for a, b, and c. a = 1 b = c = quadratic equation. x = Simplify the expression under the radical sign. x = = Evaluate the square root of 36. x = simplifying terms. First add the terms in the numerator, and then subtract them. x = The two solutions for the variable x are x = and x = 4. x = ± 4 3 ±6 +6 = 4 6 = 477. The equation is in the proper form. First, list the values for a, b, and c. a = b = 7 c = 30 = and = 4 x = 4 = 4 = adding and then subtracting in the numerator. x = = 4 4 = 6 and The two solutions for the variable x are x = 6 and x = -.5. ( 7) ± ( 7) 4()( 30) () x = ± 49 ( 40) = 4 ± () 4(1)( ) =.5 7 ± 9 ± 36 7 ±

5 47. The equation is in the proper form. First, list the values for a, b, and c. a = 6 b = 13 c = x = 1 = 1 = adding and then subtracting in the numerator. x = = = and The two solutions for the variable x are x = and x = x = = 4 1 = The equation is in the proper form. First, list the values for a, b, and c. a = 1 b = 9 c = 1 x = 36 = 36 = adding and then subtracting in the numerator. x = = = 1 6 and The two solutions for the variable x are x = 1 6 and x = 1 3. x = = First transform the equation into the proper form. Subtract from both sides of the equation. 6x + 17x = Combine like terms on both sides. 6x + 17x = 0 Now list the values for a, b, and c. a = 6 b = 17 c = x = 1 = 1 = adding and then subtracting in the numerator. x = = = and The two solutions for the variable x are x = and x = Algebra Questions x = 171 (13) ± (13) 4(6)( ) (6) 13 ± (17) ± (17) 4(6)( ) (6) 17 ± (9) ± (9) 4(1)(1) (1) 31 = ± 1 7 = 4 13 ± 41 9 ± 9 17 ± ± ± ± 31 1

6 41. First transform the equation into the proper form. Add ( 1x 3) to both sides of the equation. 14x 1x 3 = 1x + 3 1x 3 Combine like terms. 14x 1x 3 = 0 Now list the values for a, b, and c. a = 14 b = 1 c = 3 x = = = adding and then subtracting in the numerator. x = = 5 6 = and x = 1 44 = 3 = The two solutions for the variable x are x = and x = Yes, you could have divided both sides of the equation by before listing your a, b, and c values. However, the solution would have been the same. Try it yourself. 4. The equation may not appear to be in proper form because there is no value for c. But you could write it as 4x + 5x + 0 = 0, and then your values would be a = 4, b = 5, and c = 0. x = = adding and then subtracting in the numerator. x = 5 +5 = 0 = 0 and x = 5 5 = = The two solutions for the variable x are x = 0 and x = Subtract 7 from both sides of the equation. 5x 7 = 7 7 or 5x 7 = 0 In this equation, there appears to be no coefficient for the x term unless you realize that 0x = 0. So you could write the equation in the proper form like this: 5x + 0x 7 = 0 Now list the values for a, b, and c. a = 5 b = 0 c = Algebra Questions x = ± = ± = ± 35 3 = ± ± ( 1) ± ( 1) 4(14)( 3) (14) 1 ± ,79 (0) ± (0) 4(5)( 7) (5) 1 ± 1,936 (5) ± (5) 4(4)(0) (4) 5 ± ± 44 5 ± 5 5

7 Simplify the radical. The two solutions for the variable x are x = ± ± 36 = ± Transform the equation into the desired form. Subtract 1x and add 17 to both sides. 5x 1x + 17 = 1x 1x Combine like terms on both sides. 5x 1x + 17 = 0 Now list the values for a, b, and c. a = 5 b = 1 c = 17 x = = Since there is no rational number equal to the square root of a negative number, there are no solutions for this equation. 45. List the values of a, b, and c. a = 3 b = 11 c = 7 Substitute the values into the x = = = ± The solutions for the variable x are x = ± ± List the values of a, b, and c. a = 5 b = 5 c = 0 x = = = adding and then subtracting in the numerator. x = x = 5 The two solutions for the variable x are x = 0.4 and x =. ( 1) ± ( 1) 4(5)(17) (5) 1 ± (11) ± (11) 4(3)( 7) (3) (5) ± (5) 4(5)(0) (5) 5 ±, = = 0 11 ± = 0.4 and = 1 ± 16 5 ±,304 5 ± 4 67

8 47. First, multiply both sides of the equation by. x = 5x Then add 5x + to both sides of the equation. x + 5x + = 0 List the values of a, b, and c. a = b = 5 c = Substitute the values into the () = () = Find the two solutions for x by adding and then subtracting in the numerator. x = = 4 = 1 and x = = 4 The two solutions for the variable x are x = 1 and x =. 4. Transform the equation by subtracting 5 from both sides. x + x 5 = 0 List the values of a, b, and c. a = 1 b = c = 5 (5) ± (5) 4()() x = = The solution for the variable x is x = 4±1. = x = = = 4±1 49. Transform the equation by subtracting 0x and adding 19 to both sides of the equation. x 0x + 19 = 0 List the values of a, b, and c. a = 1 b = 0 c = 19 ± 4(1)( 5) (1) ± ± 4 1 ( 0) ± ( 0) 4(1)(19) (1) x = 0 ± = 0 ± adding and then subtracting in the numerator. x = = 3 = 19 and x = 0 1 = = 1 The two solutions for the variable x are x = 19 and x = 1. 5 ± 5 16 ±4 ± 1 5 ± = 0 ± Transform the equation into proper form. Use the distributive property of multiplication on the right side of the equation. 3x = 16x Subtract 16x and add to both sides. 3x 16x + = 0 List the values of a, b, and c. a = 3 b = 16 c = ( 16) ± ( 16) 4(3 (3) 6 )()

9 x = 16 ± 5 ( ) = 16 ± ( 7 3) = 16 ± 4 ( 9 3) = ± ( 3 3) = ± 3 3 = 3 3 ± 3 The solution for the variable x is x = 3 ± List the values of a, b, and c. a = 1 b = c = 11 x = ± Algebra Questions = ± 56 = ± 4 14 = ± 14 3 = 5 ± 14 The two solutions for the variable x are x = and x = List the values of a, b, and c. a = 4 b = 1 c = 6 x = 4 = 4 = adding and then subtracting in the numerator. x = = 1 4 = 1 4 and x = = 4 4 = 1 The two solutions for the variable x are x = 1 4 and x = Transform the equation into the proper form. First use the distributive property of multiplication. 7x = 4(3x) + 4(1) = 1x + 4 Then subtract (1x + 4) from both sides. 7x 1x 4 = 0 List the values of a, b, and c. a = 7 b = 1 c = 4 x = = = adding and then subtracting in the numerator. x = 1 1 x = 1 1 (1) ± (1) 4(4)( 6) (4) 1 ± ( 1) ± ( 1) 4(7)( 4) (7) 1 ± = = = 7 = and The two solutions for the variable x are x = and x = 7. () ± ( 4(1)(11) (1) 1 ± ± ± ±

10 494. You could use the fractions as values for a and b, but it might be easier to first transform the equation by multiplying it by 1. 1( 1 3 x x 3 = 0) Using the distributive property, you get 1(13x ) + 1( 3 4 x) 1(3) = 1(0). Simplify the terms. 4x + 9x 36 = 0 List the values of a, b, and c. a = 4 b = 9 c = 36 (9) ± (9) 4(4)( 36) (4) x = ± = 9± 657 = 9± 9 73 = 9± 3 73 The two solutions for the variable x are x = and x = List the values of a, b, and c. a = 5 b = 1 c = 1 x = 1 ± 14 (5 4 0 ) ± 14 ( 5) = 1 ± 4 (5 31 ) ± 31 ( 5) The two solutions for the variable x are x = and x = List the values of a, b, and c. a = 11 b = 4 c = r = 4± 1 = 4± Find the two solutions for r by adding and then subtracting in the numerator. r = 4+ 1 = = 1 and r = 4 1 = 14 = 0.64 The two solutions for the variable r are r = 1 and r = ( 1) ± ( 1) 4(5)(1) (5) ( 4) ± ( 4) 4(11)( 7) (11) 497. List the values of a, b, and c. a = 3 b = 1 c = quadratic equation. m = (1) ± (1) 4(3)( ) (3) = 6± = 4± 1 m = 1 ± = 1 ± The square root of 537 rounded to the nearest hundredth is Substitute into the expression and simplify. m = = = 0.36 and m = = = 7.36 The two solutions for the variable m are m = 0.36 and m =

11 49. Transform the equation by subtracting 16y from and adding 5 to both sides. 4y 16y + 5 = 16y 5 16y + 5 Combine like terms. 4y 16y + 5 = 0 List the values of a, b, and c. a = 4 b = 16 c = 5 quadratic equation. y = y = 16 ± 56 0 = 16 ± 176 = 16 ± = 16 ± 4 11 The square root of 11 rounded to the nearest hundredth is 3.3. Substitute into the expression and simplify. y = (3.3) = 9. = 3.66 and y = 16 4 (3.3) =. 7 = 0.34 The two solutions for the variable y are y = 3.66 and y = List the values of a, b, and c. a = 5 b = 1 c = 1 quadratic equation. s = s = = The square root of 164 rounded to the nearest hundredth is 1.1. Substitute into the expression and simplify. s = and s = 1 = 0 = = 4.1 The two solutions for the variable s are s = 0.0 and s = List the values of a, b, and c. a = 4 b = 11 c = quadratic equation. c = c = 11 ± = 11 ± 9 c = = 0.43 =.55 and c = = = 0.0 The two solutions for the variable c are c =.55 and c = List the values of a, b, and c. a = 11 b = 3 c = quadratic equation. k = 3 ± 1, ( 16) ± ( 16) 4(4)(5) (4) (1) ± (1) 4(5)( 1) (5) 1 ± ( 11) ± ( 11) 4(4)() (4) ( 3) ± ( 3) 4(11)() (11) k = = = 4.17 = ± 54 k = 3 + =.55 and k = = 7.3 = 0.36 The two solutions for the variable k are k =.55 and k = ± 164 =.4 = 11 ± ±