Solving Equations 1.0
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1 Solving Equations 1.0 MTH 18 Jane Long, Stephen F. Austin State University February 16, 2010
2 Solving Equations 1.0 Introduction: This activity is meant to give you additional practice solving equations. There will be a brief lecture during which we ll work some examples from the activity document Solving Equations 1.0, then you can finish working the rest of the problems on your own.
3 Solving Equations 1.0: Problem 1a x 2 10x = 9 x 2 10x = 27 Multiply both sides by x 2 10x + 27 = 0 Add 27 to both sides to write the problem in general form Use the Quadratic Formula: x = 10± (1)(27) 2 Check the discriminant: (1)(27) < 0, so the equation has NO (REAL) SOLUTIONS
4 Solving Equations 1.0: Problem 1a x 2 10x = 9 x 2 10x = 27 Multiply both sides by x 2 10x + 27 = 0 Add 27 to both sides to write the problem in general form Use the Quadratic Formula: x = 10± (1)(27) 2 Check the discriminant: (1)(27) < 0, so the equation has NO (REAL) SOLUTIONS
5 Solving Equations 1.0: Problem 1a x 2 10x = 9 x 2 10x = 27 Multiply both sides by x 2 10x + 27 = 0 Add 27 to both sides to write the problem in general form Use the Quadratic Formula: x = 10± (1)(27) 2 Check the discriminant: (1)(27) < 0, so the equation has NO (REAL) SOLUTIONS
6 Solving Equations 1.0: Problem 1a x 2 10x = 9 x 2 10x = 27 Multiply both sides by x 2 10x + 27 = 0 Add 27 to both sides to write the problem in general form Use the Quadratic Formula: x = 10± (1)(27) 2 Check the discriminant: (1)(27) < 0, so the equation has NO (REAL) SOLUTIONS
7 Solving Equations 1.0: Problem 1a Note: It s not necessarily important that you check the discriminant each and every time you use the Quadratic Formula, but you should know what the discriminant is and why it explains the fact that this equation has no solutions.
8 Solving Equations 1.0: Problem 1b x 8 = 2 (x 8) 1/ = 2 Rewrite the cube root as the power (this step is not essential) 1 x 8 = ( 2 ) Cube both sides to get rid of the 1 power. Cubing raises both sides to the rd power, so the power on the right hand side is ( 1 )() = 1 (which is our goal). x 8 = 27 8 Carry out the arithmetic of cubing the right side x = x = 91 8 Add 8 to both sides
9 Solving Equations 1.0: Problem 1b x 8 = 2 (x 8) 1/ = 2 Rewrite the cube root as the power (this step is not essential) 1 x 8 = ( 2 ) Cube both sides to get rid of the 1 power. Cubing raises both sides to the rd power, so the power on the right hand side is ( 1 )() = 1 (which is our goal). x 8 = 27 8 Carry out the arithmetic of cubing the right side x = x = 91 8 Add 8 to both sides
10 Solving Equations 1.0: Problem 1b x 8 = 2 (x 8) 1/ = 2 Rewrite the cube root as the power (this step is not essential) 1 x 8 = ( 2 ) Cube both sides to get rid of the 1 power. Cubing raises both sides to the rd power, so the power on the right hand side is ( 1 )() = 1 (which is our goal). x 8 = 27 8 Carry out the arithmetic of cubing the right side x = x = 91 8 Add 8 to both sides
11 Solving Equations 1.0: Problem 1b x 8 = 2 (x 8) 1/ = 2 Rewrite the cube root as the power (this step is not essential) 1 x 8 = ( 2 ) Cube both sides to get rid of the 1 power. Cubing raises both sides to the rd power, so the power on the right hand side is ( 1 )() = 1 (which is our goal). x 8 = 27 8 Carry out the arithmetic of cubing the right side x = x = 91 8 Add 8 to both sides
12 Solving Equations 1.0: Problem 1b x 8 = 2 (x 8) 1/ = 2 Rewrite the cube root as the power (this step is not essential) 1 x 8 = ( 2 ) Cube both sides to get rid of the 1 power. Cubing raises both sides to the rd power, so the power on the right hand side is ( 1 )() = 1 (which is our goal). x 8 = 27 8 Carry out the arithmetic of cubing the right side x = x = 91 8 Add 8 to both sides
13 Solving Equations 1.0: Problem 1b x 8 = 2 (x 8) 1/ = 2 Rewrite the cube root as the power (this step is not essential) 1 x 8 = ( 2 ) Cube both sides to get rid of the 1 power. Cubing raises both sides to the rd power, so the power on the right hand side is ( 1 )() = 1 (which is our goal). x 8 = 27 8 Carry out the arithmetic of cubing the right side x = x = 91 8 Add 8 to both sides
14 Solving Equations 1.0: Problem 1b Now, CHECK YOUR ANSWER! Remember that any time we raise each side of an equation to a rational power, we must check to see that we haven t introduced any extraneous solutions = 8 = 2 is a solution So 91 8
15 Solving Equations 1.0: Problem 1b Now, CHECK YOUR ANSWER! Remember that any time we raise each side of an equation to a rational power, we must check to see that we haven t introduced any extraneous solutions = 8 = 2 is a solution So 91 8
16 Solving Equations 1.0: Problem 1c t t t = 0 t (t + 4) = 0 Multiply both sides by the least common denominator t + 4 (t + 4) + 2(t + 4) = 0 Group Terms (t + 4) = 0 Factor by grouping t + 4 = 0 Divide both sides by t = 4 Subtract 4 from both sides
17 Solving Equations 1.0: Problem 1c t t t = 0 t (t + 4) = 0 Multiply both sides by the least common denominator t + 4 (t + 4) + 2(t + 4) = 0 Group Terms (t + 4) = 0 Factor by grouping t + 4 = 0 Divide both sides by t = 4 Subtract 4 from both sides
18 Solving Equations 1.0: Problem 1c t t t = 0 t (t + 4) = 0 Multiply both sides by the least common denominator t + 4 (t + 4) + 2(t + 4) = 0 Group Terms (t + 4) = 0 Factor by grouping t + 4 = 0 Divide both sides by t = 4 Subtract 4 from both sides
19 Solving Equations 1.0: Problem 1c t t t = 0 t (t + 4) = 0 Multiply both sides by the least common denominator t + 4 (t + 4) + 2(t + 4) = 0 Group Terms (t + 4) = 0 Factor by grouping t + 4 = 0 Divide both sides by t = 4 Subtract 4 from both sides
20 Solving Equations 1.0: Problem 1c t t t = 0 t (t + 4) = 0 Multiply both sides by the least common denominator t + 4 (t + 4) + 2(t + 4) = 0 Group Terms (t + 4) = 0 Factor by grouping t + 4 = 0 Divide both sides by t = 4 Subtract 4 from both sides
21 Solving Equations 1.0: Problem 1c Now, CHECK YOUR ANSWER! Remember that any time we multiply both sides of an equation by a quantity involving a variable (such as t + 4), we may introduce extraneous solutions. Try plugging in our answer above: t t t = Since the value t = 4 introduces division by 0, this value can t be a solution. So, there are NO SOLUTIONS to this equation.
22 Solving Equations 1.0: Problem 1c Now, CHECK YOUR ANSWER! Remember that any time we multiply both sides of an equation by a quantity involving a variable (such as t + 4), we may introduce extraneous solutions. Try plugging in our answer above: t t t = Since the value t = 4 introduces division by 0, this value can t be a solution. So, there are NO SOLUTIONS to this equation.
23 Solving Equations 1.0: Problem 1c Now, CHECK YOUR ANSWER! Remember that any time we multiply both sides of an equation by a quantity involving a variable (such as t + 4), we may introduce extraneous solutions. Try plugging in our answer above: t t t = Since the value t = 4 introduces division by 0, this value can t be a solution. So, there are NO SOLUTIONS to this equation.
24 Solving Equations 1.0: Problem 1c Note: The sequence of steps shown above is not the only way to solve this problem. Would you have worked the problem differently? Would you have obtained the same result?
25 Solving Equations 1.0: Problem 1f x + 2 = x x + 2 = x 2 quadratic!) Square both sides (now this looks like a 0 = x 2 x 2 Subtract x + 2 from both sides to write the equation in general form Use the Quadratic Formula: x = ± ( ) 2 4(1)( 2) 2
26 Solving Equations 1.0: Problem 1f x + 2 = x x + 2 = x 2 quadratic!) Square both sides (now this looks like a 0 = x 2 x 2 Subtract x + 2 from both sides to write the equation in general form Use the Quadratic Formula: x = ± ( ) 2 4(1)( 2) 2
27 Solving Equations 1.0: Problem 1f x + 2 = x x + 2 = x 2 quadratic!) Square both sides (now this looks like a 0 = x 2 x 2 Subtract x + 2 from both sides to write the equation in general form Use the Quadratic Formula: x = ± ( ) 2 4(1)( 2) 2
28 Solving Equations 1.0: Problem 1f So x = ± Now, let s think about how we should check our answer. The possible solutions we obtained are not easily plugged in, but we can plug the values into our calculator in order to verify. Since these values represent all of the possible solutions to the equation, we can be satisfied that these values are, in fact, solutions as long as the left hand side x + 2 agrees with the right hand side x up to one decimal or so. Check these values yourself to decide.
29 Solving Equations 1.0: Practice On Your Own Now, you are ready to work through the rest of the activity on your own. Discuss the problems with your fellow students and/or your instructor if you need help!
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