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QUIZ 3 – detailed solutions
1) Solve (x 2 ‐ 1) / x + 1 = 0 Begin by observing x not = ‐1, since x = ‐1 would cause the left member to be undefined. Next, if the fraction equals 0 then the numerator must be 0. So we solve x 2 ‐ 1 = 0 Factoring (x + 1)(x ‐ 1) = 0. Setting each factor equal to 0 x + 1 = 0 and x ‐ 1 = 0 Solving x = ‐1 and x = 1 We reject x = ‐1 for the reasons listed above. Thus, the only answer is x = 1. 2) Solve 3 / (a+b) = c for a. Multiplying by a + b we get 3 = c(a + b) Applying the distributive property 3 = ac + bc subtracting 3 ‐ bc = ac dividing by c (3 ‐ bc) / c = a or 3/c ‐ bc/c = a or 3/c ‐ b = a. 3) Solve 2/x = x + 1 Begin by observing x not = 0. Next, multiply by x and obtain 2 = x(x + 1) Distributing 2 = x 2 + x Since the equation is quadratic we get one side to 0 0 = x 2 + x ‐ 2 Factoring 0 = (x + 2)(x ‐ 1) Setting the factors to 0 x + 2 = 0 and x ‐ 1 = 0 Solving x = ‐2 and x = 1. 4) Solve sqrt{6x + 7} = x + 2 Eliminate the square root by squaring both sides sqrt{ 6x + 7} 2 = (x + 2) 2 which is 6x + 7 = (x + 2)(x + 2) or 6x + 7 = x 2 + 4x + 4 getting one side to 0 0 = x 2 ‐ 2x ‐ 3 factoring 0 = (x ‐ 3)(x + 1) setting factors to 0 0 = x ‐ 3 and 0 = x + 1 solving 3 = x and ‐1 = x both of these answers check. 5) Solve A =  r 2 for r Divide by  on both sides
A/ = r 2
Performing the square root on both sides and remembering + and ‐ values must be considered +/‐ (A /  ) 1/2 = r
6) sqrt{2x + 1} + 1 = x Isolating the square root sqrt{2x + 1} = x‐1 Eliminating the square root by squaring both sides yields sqrt{2x+1} 2 = (x‐1) 2 which is 2x + 1 = (x ‐ 1)(x ‐ 1) or 2x + 1 = x 2 ‐ 2x + 1 Getting one side to 0 0 = x 2 ‐ 2x + 1 ‐ 2x ‐ 1 or 0 = x 2 ‐ 4x Factoring 0 = x(x ‐ 4) Setting the individual factors to 0. 0 = x and 0 = x ‐ 4 or 0 = x and 4 = x. However, checking x = 0 we see sqrt{ 2(0) + 1} + 1 = 0 sqrt{1} + 1 = 0 1 + 1 = 0 2 = 0 So the only solution is x = 4. 7) Solve x 2 ‐ x ‐ 1 = 0 Since the equation does not factor we employ the quadratic formula. x = (‐ b +/‐ sqrt{b 2 ‐ 4a} ) / 2a where a = 1 b = ‐1 and c = ‐1 x = (1 +/‐ sqrt{ 1 ‐ 4(1)(‐1) } ) / 2(1) x = (1 +/‐ sqrt{1 + 4} ) / 2 x = (1 +/‐ sqrt{5} )/ 2 8) Solve 2x 3 + 3x 2 = 7x – 2 Getting one side to 0 2x 3 + 3x 2 ‐ 7x + 2 = 0 By observation x = 1 is a solution, thus (x – 1) must be a factor. That is; (x – 1)( ? ? ? ) = 2x 3 + 3x 2 ‐ 7x + 2 = 0 Now, it is clear that the first and last terms of the unknown factor must be 2x 2 and ‐2, respectively. That is, (x – 1)( 2x 2 ? ‐2 ) = 2x 3 + 3x 2 ‐ 7x + 2 = 0 Now, using some educated guessing and checking we come to the conclusion that the unknown middle factor must be + 5x. So (x – 1)( 2x 2 + 5x ‐2 ) = 2x 3 + 3x 2 ‐ 7x + 2 = 0 Next, clearly one solution for the equation is x = 1 . And since 2x 2 + 5x ‐2 does not factor we employ the quadratic formula (x = ‐b +/‐ sqrt{b 2 ‐ 4ac} / 2a) with a = 2 b = 5 and c = ‐2. Plugging in these values x = ‐5 +/‐ sqrt{ 25 – 4(2)(‐2)} / 2(2) or x = ‐5 +/‐ sqrt{25 + 16} / 4 so x = ‐5 +/‐ sqrt{41} / 4 giving us two more solutions. 9) Solve x + 5y = 11 x + y = 3 Subtracting the bottom equation from the top 4y = 8 and y = 2 Consequently, since y = 2 we can use either of the optional equations and get x + 2 = 3 so x = 1 and finally x = 1, y = 2 checks in the other original equation x + 5y = 11 1 + 5(2) = 11 1 + 10 = 11 11 = 11 10) Solve 2y + z = ‐2 3x + 5y ‐ 5z = 1 2x + 4y ‐ 2z = 2 Many strategies can be employed to solve problems of this type. We begin by solving the first equation for z z = ‐2 ‐ 2y substituting it into both the second and third equations 3x + 5y ‐ 5(‐2 ‐ 2y) = 1 2x + 4y ‐ 2(‐2 ‐ 2y) = 2 using the distribution property 3x + 5y + 10 + 10y = 1 2x + 4y + 4 + 4y = 2 collecting terms 3x + 15y = ‐9 2x + 8y = ‐2 solving simultaneously 6x + 30y = ‐18 6x + 24y = ‐6 subtracting 6y = ‐12 and y = ‐2 Substituting y = ‐2 into the first equation 2y + z = ‐2 2(‐2) + z = ‐2 ‐4 + z = ‐2 z = 2 Substituting y = ‐2 and z = 2 into equation 2 we get 3x + 5y ‐ 5z = 1 3x + 5(‐2) ‐ 5(2) = 1 3x ‐ 10 ‐ 10 = 1 3x = ‐21 x = ‐7 Giving final values of x = 7 y = ‐2 z = 2 which can be checked in the third equation. Brad Smith CUmath Clarkson University Potsdam, NY 13699