3.5 Solving Quadratic Equations by the

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1 Chapter 3. Quadratic Equations and Quadratic Functions 3.5 Solving Quadratic Equations y the Quadratic Formula Learning ojectives Solve quadratic equations using the quadratic formula. Identify and choose methods for solving quadratic equations. Solve real-world prolems using functions y completing the square. Introduction In this section, you will solve quadratic equations using the Quadratic Formula. Most of you are already familiar with this formula from previous mathematics courses. It is proaly the most used method for solving quadratic equations. For a quadratic equation in standard form The solutions are found using the following formula. ax + x + c = 0 x = ± 4ac a We will start y explaining where this formula comes from and then show how it is applied. This formula is derived y solving a general quadratic equation using the method of completing the square that you learned in the previous section. Divide y the coefficient of the x term: x + a x = c ( ) a Rewrite: x + x = c a a ( ) ( ) ( ) Add the constant to oth sides: x + x + = c a a a a + 4a ( Factor the perfect square trinomial: x + ) = 4ac a 4a + 4a ( Simplify: x + ) = 4ac a 4a Take the square root of oth sides: x + a = 4ac 4a and x + a = 4ac 4a Simplify: x + a = 4ac and x + a a = 4ac a x = a + 4ac and x = a a 4ac a x = + 4ac and x = 4ac a a 185

2 3.5. Solving Quadratic Equations y the Quadratic Formula This can e written more compactly as x = ± 4ac a. You can see that the familiar formula comes directly from applying the method of completing the square. Applying the method of completing the square to solve quadratic equations can e tedious. The quadratic formula is a more straightforward way of finding the solutions. Solve Quadratic Equations Using the Quadratic Formula Applying the quadratic formula asically amounts to plugging the values of a, and c into the quadratic formula. Example 1 Solve the following quadratic equation using the quadratic formula. a) x + 3x + 1 = 0 ) x 6x + 5 = 0 c) 4x + x + 1 = 0 Solution Start with the quadratic formula and plug in the values of a, and c. a) Quadratic formula x = ± 4ac a Plug in the values a =, = 3,c = 1. x = 3 ± (3) 4()(1) () Simplify. x = 3 ± Separate the two options. x = Solve. = 3 ± 1 4 and x = x = 4 = 1 4 and x = 4 = 1 Answer x = 1 and x = 1 Rememer you can check this soltuion y determine the x-intercepts of the quadratic function y = x + 3x + 1 ) Answer x = 5 and x = Quadratic formula. x = ± 4ac a Plug in the values a = 1, = 6,c = 5. x = ( 6) ± ( 6) 4(1)(5) (1) Simplify. x = 6 ± 36 0 Separate the two options. x = and x = 6 4 = 6 ± 16 Solve x = 10 = 5 and x = = 1

3 Chapter 3. Quadratic Equations and Quadratic Functions c) Quadratic formula. Plug in the values a = 4, = 1,c = 1. x = ± 4ac a x = 1 ± (1) 4( 4)(1) ( 4) Simplify. x = 1 ± Separate the two options. x = Solve. x.39 and x.64 = 1 ± 17 8 and x = Answer x.39 and x.64 Often when we plug the values of the coefficients into the quadratic formula, we otain a negative numer inside the square root. Since the square root of a negative numer does not give real answers, we say that the equation has no real solutions. In more advanced mathematics classes, you will learn how to work with complex (or imaginary ) solutions to quadratic equations. Example Solve the following quadratic equation using the quadratic formula x + x + 7 = 0 Solution: a) Quadratic formula. x = ± 4ac a Plug in the values a = 1, =,c = 7. x = ± () 4(1)(7) (1) Simplify. x = ± 4 8 = ± 4 Answer There are no real solutions. To apply the quadratic formula, we must make sure that the equation is written in standard form. For some prolems, we must rewrite the equation efore we apply the quadratic formula. Example 3 Solve the following quadratic equation using the quadratic formula. a) x 6x = 10 ) 8x + 5x = 6 Solution: 187

4 3.5. Solving Quadratic Equations y the Quadratic Formula a) Rewrite the equation in standard form. x 6x 10 = 0 Quadratic formula Plug in the values a = 1, = 6,c = 10. x = ± 4ac a x = ( 6) ± ( 6) 4(1)( 10) (1) Simplify. x = 6 ± = 6 ± 76 Separate the two options. x = and x = 6 76 Solve. x 7.36 and x 1.36 Answer x 7.36 and x We should give the exact solution unless we are told to round. x = 6 76 ) Rewrite the equation in standard form. 8x + 5x + 6 = 0 Quadratic formula x = ± 4ac a Plug in the values a = 8, = 5,c = 6. x = 5 ± (5) 4(8)(6) (8) Simplify. x = 5 ± = 5 ± = 5 ± 167i 16 Answer No real solutions, two complex solutions: 5± 167i 16 Notice if we try to check this solution y graphing the quadratic function y = 8x + 5x + 6, the graph does not cross the x-axis or have x-intercepts. This verifies we have complex solutions with an imaginary part. Finding the Vertex of a Paraola with the Quadratic Formula Sometimes you get more information from a formula eyond what you were originally seeking. In this case, the quadratic formula also gives us an easy way to locate the vertex of a paraola. First, recall that the quadratic formula tells us the roots or solutions of the equation ax + x + c = 0. Those roots are x = ± 4ac. a 188

5 Chapter 3. Quadratic Equations and Quadratic Functions We can rewrite the fraction in the quadratic formula as x = a ± 4ac. a Recall that the roots are symmetric aout the vertex. In the form aove, we can see that the roots of a quadratic equation are symmetric around the x coordinate a ecause they move 4ac a units to the left and right (recall the ± sign) from the vertical line x = a. The image to the right illustrates this for the equation x x 3 = 0. The roots, -1 and 3 are oth units from the vertical line x = 1. Identify and Choose Methods for Solving Quadratic Equations. In mathematics, you will need to solve quadratic equations that descrie application prolems or that are part of more complicated prolems. You learned four ways of solving a quadratic equation. Factoring. Taking the square root. Completing the square. Quadratic formula. Usually you will not e told which method to use. You will have to make that decision yourself. However, here are some guidelines to which methods are etter in different situations. Factoring is always est if the quadratic expression is easily factorale. It is always worthwhile to check if you can factor ecause this is the fastest method. Taking the square root is est used when there is no x term in the equation. Completing the square can e used to solve any quadratic equation. This is usually not any etter than using the quadratic formula (in terms of difficult computations), however it is a very important method for re-writing a quadratic function in vertex form. It is also used to re-write the equations of circles, ellipses and hyperolas in standard form (something you will do in algera II, trigonometry, physics, calculus, and eyond...). Quadratic formula is the method that is used most often for solving a quadratic equation if solving directly y taking square root and factoring does not work. If you are using factoring or the quadratic formula make sure that the equation is in standard form. Example 4 Solve each quadratic equation a) x 4x 5 = 0 ) x = 8 c) 4x + x = d) 5x 9 = 0 e) 3x = 8x Solution a) This expression if easily factorale so we can factor and apply the zero-product property: 189

6 3.5. Solving Quadratic Equations y the Quadratic Formula Factor. (x 5)(x + 1)=0 Apply zero-product property. x 5 = 0 and x + 1 = 0 Solve. x = 5 and x = 1 Answer x = 5 and x = 1 ) Since the expression is missing the x term we can take the square root: Take the square root of oth sides. x = 8 and x = 8 Answer x =.83 and x =.83 c) Rewrite the equation in standard form. It is not apparent right away if the expression is factorale, so we will use the quadratic formula. Quadratic formula Plug in the values a = 4, = 1,c =. x = ± 4ac a x = 1 ± 1 4( 4)( ) ( 4) Simplify. x = 1 ± = 1 ± 31 8 = 1 ± 31i 8 Answer Two complex solutions: x = 1± 31i 8 = 1 8 ± 31 8 i d) This prolem can e solved easily either with factoring or taking the square root. Let s take the square root in this case. Add 9 to oth sides of the equation. 5x = 9 Divide oth sides y 5. x = 9 5 Take the square root of oth sides. x = 9 5 and x = Simplify. x = 3 5 and x = Answer x = 3 5 and x = 3 5 e) Rewrite the equation in standard form 3x 8x = 0 Factor out common x term. x(3x 8)=0 Set oth terms to zero. x = 0 and 3x = 8 Solve. x = 0 and x = 8 3 Answer x = 0 and x =

7 Chapter 3. Quadratic Equations and Quadratic Functions Solve Real-World Prolems Using Quadratic Functions y any Method Here are some application prolems that arise from numer relationships and geometry applications. Example 5 The product of two positive consecutive integers is 156. Find the integers. Solution For two consecutive integers, one integer is one more than the other one. Define Let x = the smaller integer x + 1 = the next integer Translate The product of the two numers is 156. We can write the equation: Solve x(x + 1)=156 Apply the quadratic formula with a = 1, = 1, c = 156 x + x = 156 x + x 156 = 0 x = 1 ± 1 4(1)( 156) (1) x = 1 ± 65 1 ± 5 = x = x = and x = 1 5 = 1 and x = 6 = 13 Since we are looking for positive integers take, x = 1 Answer 1 and 13 Check 1 13 = 156. The answer checks out. Example 6 The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875 square/meters. Find the dimensions of the pool. Solution: 191

8 3.5. Solving Quadratic Equations y the Quadratic Formula Draw a sketch Define Let x = the width of the pool x + 10 = the length of the pool Translate The area of a rectangle is A = length width, so x(x + 10)=875 Solve x + 10x = 875 x + 10x 875 = 0 Apply the quadratic formula with a = 1, = 10 and c = 875 x = 10 ± (10) 4(1)( 875) (1) x = 10 ± x = 10 ± ± 60 = x = and x = x = = 5 and x = = 35 Since the dimensions of the pools should e positive, then x = 5 meters. Answer The pool is 5 meters 35 meters. Check 5 35 = 875 m. The answer checks out. Example 7 Suzie wants to uild a garden that has three separate rectangular sections. She wants to fence around the whole garden and etween each section as shown. The plot is twice as long as it is wide and the total area is 00 ft. How much fencing does Suzie need? 19

9 Chapter 3. Quadratic Equations and Quadratic Functions Solution Draw a Sketch Define Let x = the width of the plot x = the length of the plot Translate Area of a rectangle is A = length width, so x(x)=00 Solve x = 00 Solve y taking the square root. x = 100 x = 100 and x = 100 x = 10 and x = 10 We take x = 10 since only positive dimensions make sense. The plot of land is 10 feet 0 feet. To fence the garden the way Suzie wants, we need lengths and 4 widths = (0)+4(10)=80 feet of fence. Answer: The fence is 80 feet. Check 10 0 = 00 ft and (0)+4(10)=80 feet. The answer checks out. Example 8 An isosceles triangle is enclosed in a square so that its ase coincides with one of the sides of the square and the tip of the triangle touches the opposite side of the square. If the area of the triangle is 0 in what is the area of the square? Solution: Draw a sketch. 193

10 3.5. Solving Quadratic Equations y the Quadratic Formula Define Let x = ase of the triangle x = height of the triangle Translate Area of a triangle is 1 ase height, so 1 x x = 0 Solve 1 x = 0 Solve y taking the square root. x = 40 x = 40 = 10 and x = 40 = 10 x 6.3 and x 6.3 The side of the square is approximately 6.3 inches. The area of the square is (6.3) 40 in, twice as ig as the area of the triangle. Answer: Area of the triangle is 40 in Check: It makes sense that the area of the square will e twice that of the triangle. If you look at the figure you can see that you can fit two triangles inside the square. 194

11 Chapter 3. Quadratic Equations and Quadratic Functions FIGURE 3. The answer checks. Review Questions Solve the following quadratic equations using the quadratic formula. 1. x + 4x 1 = 0. x 6x = x 1 x = x + x 3 = 0 5. x 7x + 1 = x + 5x = x = 0 8. x + x + 6 = 0 Solve the following quadratic equations using the method of your choice. 9. x x = x 1 = x + 5x 3 = 0 1. x + 7x 18 = x + 6x = x x = x + 1x + 1 = x + 6x + 9 = 0 195

12 3.5. Solving Quadratic Equations y the Quadratic Formula x + 1 = x + 4x = x 1 = 0 0. x x 3 = 0 1. The product of two consecutive integers is 7. Find the two numers.. The product of two consecutive odd integers is 1 less than 3 times their sum. Find the integers. 3. The length of a rectangle exceeds its width y 3 inches. The area of the rectangle is 70 square inches, find its dimensions. 4. Angel wants to cut off a square piece from the corner of a rectangular piece of plywood. The larger piece of wood is 4 feet 8 feet and the cut off part is 1 3 of the total area of the plywood sheet. What is the length of the side of the square? 5. Mike wants to fence three sides of a rectangular patio that is adjacent the ack of his house. The area of the patio is 19 ft and the length is 4 feet longer than the width. Find how much fencing Mike will need. Review Answers 1. x = 7,x = 3. x = 3 ± 1 3. x = 1 1 ± x = 3,x = 1 5. x = 7 ± x = 0,x = x = ± i 5 9. x =,x = x = ± x = 1,x = 3 1. x = 9,x = 13. x = 1 ± i x = 0,x = x = ± x = x = ± 1 9 i ± i 19. x = ± x = 1,x = and 9 196

13 Chapter 3. Quadratic Equations and Quadratic Functions. 5 and in and 10 in 4. side = 3.7 ft feet of fencing. 197