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Minis March 2016 Activity Solutions Warm-Up! 1. Solve the equation √(x + 4) = 3 by squaring both sides. We get x +4 = 9, so x = 5. 2. When we expand the given product we get (a + b)(a + b) = a2 + ab + ab + b2 = a2 +2ab + b2. 3. If we subtract the second equation from the first equation we get u + v + w + x + y + z = 45 − ( v + w + x + y + z = 21 ) u = 24 4. Let x represent the first number and y represent the second number. We are told that x + y = 6 and xy = 7. We are asked to find the sum of the reciprocals of the two numbers, 1/x + 1/y, which can be rewritten as (y + x)/xy. Substituting, we have (y + x)/xy = 6/7. The Problems are solved in the Minisvideo. Follow-up Problems 5. Since we ultimately want the product of the three integers, let’s start by multiplying the products of the pairs we are given. We have (xy)(yz)(xz) = 4 × 18 × 50 → x2y2z2 = 3600 → (xyz)2 = 3600. Now if we take the square root of each side, we get xyz = 60, since x, y and z are positive numbers. 6. We are told that xyz = 45 and 1/x + 1/y + 1/z = 1/5. We can rewrite the left side of the second equation using a common denominator to get (yz/xyz) + (xz/xyz) + (xy/xyz) = 1/5 → (xy + xz + yz)/xyz = 1/5. But we know that xyz = 45, so we have (xy + xz + yz)/45 = 1/5 → xy + xz + yz = 9. If the sum of the three products xy, xz and yz is 9, then their mean is 9/3 = 3. 7. Let’s start by cubing each side of the given equation to get ( ) ( )( ) )( ) ( a+ a+ 1 a 1 a 2 a+ 3 = 33 1 = 27 a 1 1 = 27 a+ a2 a 2 1 1 a 3 + a + 2a + + + 3 = 27 a a a 3 1 a3 + 3a + + 3 = 27. a a 1 1 a3 + 3 a + + 3 = 27. a a © 2016 MATHCOUNTS Foundation. All rights reserved. MATHCOUNTS Mini Solution Set 1 a 3 + 3 ( 3 ) + 3 = 27 a 1 = 27 a3 + 9 + a2 + 2 + ( ) ( ( ( ( a )) ) ) )( ( )) ( )( aa 1 2 1 a 2 + 2a ++ 1 2 a + 27 = 27 a aa 2 1 11 1 a 3 + aa 2++2a = 27 27 2 ++ a 2+ aa ++ a 3 = a a 3 1 1 two middle terms of the expression on the left-hand side of the equation, If we factor a 3 out of the 2 1 a3 + 27. a 3 + a + 2a + 3a ++ a + + a 33 = = 27 we have a a a 1 1 3 a a+33+ 3 a a++ 3 + + 13 = = 27. aa aa 3 27. 1 1 =127 3 a33 a+ + 3 (13 ) + +a + 3 = 27. Since awe+know that a a a 3 a = 3, we can substitute and simplify to get 1 3 1 = 27 27 a 3 +a3 (+39) + + a 33 = a 1 3 + 13 = a 3 +a9 + = 18. 27 aa 3 1 a 3 + 3 = 18. a ( ) ( ) 8. First, let's multiply the two given equations together ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) 1 1 1 y+ =2 2 z y 1 1 1 x + x y + 1= 2 xy +y1 + 1 +z1 = 1 2 1 x + z y + yz= 2 2 y x x z 11 xy +xy ++ 1++ ==0.1 zx z yz yz 1 + xy + + 1 1 =1 z + x + 1yz 0. xy ==0. xyz + x + zx z,yyz Multiplying through by 1we get the equation = 0. xy + + 1 zx + + 2yz==00. xyz +xyz y1 + ==−2. 0. xyz + x xyz +2 Substituting in andxyz solving y =for0 xyz, we find that x+ + 2 ==−2. 0 xyz xyz xyz = −2. © 2016 MATHCOUNTS Foundation. All rights reserved. MATHCOUNTS Mini Solution Set