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What two numbers whose product is 4 and the sum of whose reciprocals is 65/56.

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3 Answers

xy = 4 => y = 4/x => 1/y = x/4
 
1/x + 1/y = 65/56
 
1/x + x/4 = 65/56
 
Multiply both sides by 56x:
 
56 + 14x^2 = 65x
 
14x^2 - 65x + 56 = 0
 
[I chose to find Vertex Form and solve that.]
 
h = -b/2a = - -65/(2*14) = 65/28
 
k = c - ah^2 = 56 - 14(65/28)^2
= (56^2 - 65^2)/56 = (56-65)(56+65)/56 = -9(121)/56
 
x = h ± √(-k/a) = 65/28 ± √(- -9(121)/(56*14))
 
= 65/28 ± √(3^2(11^2)/(4*14^2))
 
= 65/28 ± 3*11/(2*14) = (65 ± 33)/28
 
x = 7/2 or 8/7
 
y = 4/x = 8/7 or 7/2
 
So the two numbers are 8/7 and 7/2.
 
check:
 
xy =? 4
(8/7)(7/2) =? 4
4 = 4   √
 
1/x + 1/y =? 65/56
1/(8/7) + 1/(7/2) =? 65/56
7/8 + 2/7 =? 65/56
7*7/(8*7) + 2*8/(7*8) =? 65/56
(49 + 16)/56 =? 65/56
65/56 = 65/56  √
 XY = 4
 
 1/ X + 1/ Y = 65/ 56
 
 
   X = 4 / Y 
 
    Substitute in 2nd equation:
 
       Y/ 4 +  1 / Y = 65 / 56
 
 
      Y^2 + 4 = 65/ 56
        4 Y 
      
       56 Y ^2 + 224 = 260 Y
 
        14 Y ^2 - 65 Y - 56 = 0
    
       14Y^2 - 16 Y - 49 Y - 56 =0
 
        2 Y ( 7 Y - 8 ) - 7 (  7Y - 8 ) =0
 
         ( 7Y - 8 ) ( 2Y - 7 ) = 0
 
        7y - 8 = 0     Y = 8/7      X = 4 / ( 8/7) = 7/2
 
          Test:
 
             7/8 + 2 /7 =  ( 49 + 16) /56 = 65/ 56
 
              True:
 
        2Y - 7 = 0      Y = 7/2      X = 8/7
 
                    This will work too, only X, y change their position.
 
           
 
 
Hi Willie,
 
Let twonumbers are x & y.
x*y=4...................Eq-1
1/x+1/y=65/56.....Eq-2
Now from Eq-1, y=4/x
substituting value of y in Eq-2,
we get,1/x+1/4/x=65/56
           1/x+x/4=65/56
            4+x^2=65/56*4x
            x^2+4=65/14 *x
            x^2-65x/14+4=0
           14x^2-65x+56=0
           14x^2-49x-16x+56=0
           7x(2x-7)-8(2x-7)=0
            (2x-7)(7x-8)=0
            2x-7=0  or 7x-8=0
             x=7/2 or x=8/7
             y=4/x=4/7/2=8/7  or  y=4/8/7=7/2.
Ans: two numbers are 7/2 & 8/7.
 
Meena from Strongsville,OH

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