# MATH 108 REVIEW TOPIC 10 Quadratic Equations. B. Solving Quadratics by Completing the Square

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1 Math 108 T10-Review Topic 10 Page 1 MATH 108 REVIEW TOPIC 10 Quadratic Equations I. Finding Roots of a Quadratic Equation A. Factoring B. Quadratic Formula C. Taking Roots II. III. Guidelines for Finding Roots of a Quadratic Completing the Square A. Perfect Square Trinomials B. Solving Quadratics by Completing the Square Answers to Exercises

3 Math 108 T10-Review Topic 10 Page 3 Solution: x(x ) = 5 x x 5=0 (x 5)(x +1)=0 ab =0 x =5orx = 1 Example: Solve 9x = x. Solution: 9x = x 9x x =0 x(9x 1) = 0 ab =0 x =0orx = 1 9 What if you attempted the same problem using the following method? 9x = x 9x = 1 divide by x x = 1 9 Every quadratic equation has roots. Dividing by x removes the root x =0. However, dividing by a constant does not effect roots. Example: Solve 16t +80t =0. 16t +80t =0 t 5t = 0 t(t 5) = 0 t =0ort =5 divide by ( 16) Here s one last example of how factoring finds roots. Example: x 3 +3x x 1 = 0. Even though the example is not quadratic, any factorable polynomial can be solved using the same principles. Solution: x 3 +3x x 1 = 0 x (x +3) (x +3)=0 (x )(x +3)=0 ] Grouping Review Topic (x )(x + )(x +3)=0 x = ± or x = 3

4 Math 108 T10-Review Topic 10 Page Exercise 1: Solve for x. a) x(x +1)=15 b) 1x +60x +75=0 c) 5x x + 3 x += d) x 3 9x =0 6 x x Hint: Find LCD and clear fractions. e) x 3 5x 18x +5=0 Answers B. Quadratic Formula Another method for finding roots to a quadratic equation is the quadratic formula. For ax + bx + c =0,a 0, x = b ± b ac a Example: Solve x 6x =0. Solution: With a =1,b= 6 andc =,. 5 = 13 = 13 x = 6 ± ( 6) (1)( ) (1) = 6 ± 5 =3± 13 = 6 ± 13 Any help you need with simplifying radicals can be found in Review Topic 6. Comments: A quadratic has real roots when b ac 0. The discussion of non-real or complex roots (when b ac < 0) will be left for the course.

5 Math 108 T10-Review Topic 10 Page 5 C. Taking Roots For quadratics with no middle term (when b = 0), the simplest approach is to take square roots of both sides of the equation. Example: Solve 3x =0. Solution: 3x =0 x = isolate x and take roots. 3 x = ± = ± Common error: Don t forget the negative root. Example: Solve (x ) =17. x =5 x = 5 or 5. You could expand the binomial, put into quadratic form and solve. Instead, as in the previous example, square root both sides of the equation. Solution: (x ) =17 x =± 17 x =± 17 It will help if you keep this example in mind when looking at Section III of this topic, completing the square. Exercise : Find all real solutions. a) x x = 0 e) 3x + = x x 3 b) (x +3) = f) 5 3 x +3x +1=0 c) x x 3=0 g) k = 1 mv for v (Kinetic energy) d) x 3x +=0 Answers

6 Math 108 T10-Review Topic 10 Page 6 II. Guidelines for Finding Roots of a Quadratic You should now be able to solve quadratic equations using any of the three methods shown: factoring, quadratic formula, or taking roots. Here is a summary of what has been covered. 1) For ax + c = 0, isolate x and square root both sides. Don t forget the negative root. Otherwise... ) Put into the form ax + bx + c =0. This may require removing parentheses or clearing fractions. Dividing out a constant is helpful but not necessary. 3) Find roots by factoring or the *quadratic formula. If b ac < 0, the equation has no real roots. ) Check solutions, especially if original equation is fractional. *Don t overuse the quadratic formula. Factoring is an important skill to maintain so use it at every opportunity. III. Completing the Square Section Ic) demonstrated how quadratics in the form (x ± ( solved. )) = k are Illustration: (x ) =17 x =± 17 x =± 17. How is this related to completing the square? By expanding (x ) and setting equal to 0, (x ) =17 x x 13 = 0. This would seem to indicate that any quadratic can be changed into (x ± ( )) = k form (and then solved). Such a process is called completing the square. A. Perfect Square Trinomials Completing the square requires a thorough understanding of how trinomials of the form a ± ab + b always factor into (a ± b) (Review Topic ).

7 Math 108 T10-Review Topic 10 Page 7 Illustration: a +ab+b (a+b) { }} { { }} { x +6x +9= x + (3)x +3 = (x +3) x 10x +5=x (5)x +5 =(x 5) x 5x + 5 ( ) ( ) ( 5 5 = x x + = x 5 Trinomials such as these are referred to as Perfect Square Trinomials (PST). Exercise 3: Find the term needed to make a PST, then express in factored form. ( ) ( ) ( ) ( x 5x +? = x x +? = x x + = x 5 ) ) a) x 1x +? b) a +9a +? c) x 1 x +? Answers B. Solving Quadratics by Completing the Square Example: Solve x 10x =0. Solution. x 10x = x (5)x +5 =+5 (x 5) =9 x 5=± 9 x =5± 9 Move the constant so it won t interfere with completing the square Comp. Square, maintain equality by adding to both sides Factor and take roots

8 Math 108 T10-Review Topic 10 Page 8 Example: Solve x +7x 15 = 0 Solution. x + 7 x = 15 ( 7 x + ( x + 7 ) x + ) = ( 7 ) = 15 + x + 7 = ±13 x = 7 ± 13 = 3 or 5 ( ) 7 Why are these the roots of x +7x 15 = 0? Divide out coef. of x Check: if x = 3 ( ( ) 3 3 ), = =0 if x = 5, ( 5) +7( 5) 15 = = 0. Final Comment: Completing the square may not be your preferred method for solving quadratics. However, the process is important to learn. You will need to complete squares when working with equations of circles and parabolas. Exercise : Solve by completing the square. a) x +8x 6=0 b) x 11x =0 c) 5 + x x =0 Answers Beginning of Topic 108 Skills Assessment

9 Math 108 Exercise 1 Topic 10 Page 9 Solve for x. a) x(x + 1) = 15 (d) x 3 9x =0 b) 1x +60x + 75 = 0 (e) x 3 5x 18x +5=0 c) 5x x + 3 x += 6 x x Hint: Find LCD and clear fractions. Answers: a) x + x 15 = 0 (x 5)(x +3)=0 x = 5 or x = 3 b) 1x +60x +75=0 x +0x +5=0 (x +5) =0 x = 5 Since (x + 5) is a repeated factor, 5 is a repeated or double root. c) Multiply by x(x ) to clear fractions. 5x +3(x ) + x(x ) = 0 7x x =0 x(7x 1) = 0 Thus x = 0 or x = 1 7 appear to be roots. Since division by 0 is not permissible, the only root is 1. Always check roots when variables 7 appear in any denominator. d) x(x 9) = 0 x(x 3)(x +3)=0 x =0 or ± 3 e) x 3 5x 19x +5=0 x (x 5) 9(x 5) = 0 (x 9)(x 5) = 0 x = ±3 or 5 Return to Review Topic

10 Math 108 Exercise Topic 10 Page 10 Find all real solutions. a) x 10 = 0 e) x +1 3x + = x x 3 b) (x +3) = f) 5 3 x +3x +1=0 c) x x 3=0 g) k = 1 mu for v (Kinetic energy) d) x 3x +=0 Answers: a) x =5 x = 5or 5 b) x +3=± x = 3 ± x = 1 orx = 5 c) x = ± 7 = 1 ( ± 3 3) d) x = 3 ± 7 ; Since b ac < 0, this equation has no real roots. e) Solving as a proportion, (x 3)(x +1)=(3x + )(x ) x 3x 1=0 x = 3 ± 13 f) After clearing fractions, 5x +9x +3=0. x = 9 ± 1 = 1 ( 9 ± 1) g) k = 1 mv k = mv v = k m k v = m Since we solved for velocity, disregard the negative root. Return to Review Topic

11 Math 108 Exercise 3 Topic 10 Page 11 Find the term needed to make a PST, then express in factored form. a) x 1x +? b) a +9a +? c) x 1 x +? Answers: a) x (7)x +7 =(x 7) b) a + c) x ( ) 9 a + ( ) 1 x + ( ) ( 9 = a + 9 ) ( ) ( 1 = x 1 ) Return to Review Topic

12 Math 108 Exercise Topic 10 Page 1 Solve by completing the square. a) x +8x 6=0 b) x 11x =0 c) 5 + x x =0 Answers: a) (x +) =6+ x +=± x = ± b) x x 11 x =0 ( ) ( ) ( x + = ( x 11 ) ( 11 = ) ) x 11 = ±11 x = 11 ± 11 =0or 11 c) x x =5 (x ) =9 x =± 3=5or 1 Return to Review Topic

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