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Three consecutive, odd, positive integers have the property that the product of the smaller two is 3 more than four times the sum of the larger two. Algebraically determine the three integers.

 Apr 30, 2020
 #1
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Let  x,  x + 2,  and  x + 4  represent the three consecutive positive integers.

 

The product of the smaller two:  x(x + 2)

3 more than four times the sum of the larger two:  3 + 4[ (x + 2) + (x + 4) ]

 

       x(x + 2)  =  3 + 4[ (x + 2) + (x + 4) ]

        x2 + 2x  =  3 + 4[ 2x + 6 ]

        x2 + 2x  =  3 + 8x + 24

        x2 + 2x  = 27 + 8x

  x2 - 6x - 27  =  0

(x - 9)(x + 3)  =  0

x  =  9

x + 2  =  11

x + 4  =  13

 Apr 30, 2020
 #2
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The numbers are x, x+2, and x+4 

 

The equation is                   (x)(x+2) = 3 + (4)[(x+2)+(x+4)] 

 

                                            x2 + 2x = 3 + (4)(2x+6)  

 

                                             x2 + 2x = 3 + 8x + 24 

 

rearrange                            x2 + 2x – 3 – 8x – 24 = 0 

 

                                            x2 – 6x – 27 = 0 

 

factor                                   (x – 9)(x + 3) = 0 

 

                                             x = 9  and x =–3  but discard –3 because the problem calls for positive numbers 

 

                                             The numbers are 9, 11, and 13 

.

 Apr 30, 2020

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