Given that a , b , c ∈ Z a &gt; 10and ( x − a ) ( x − 12 ) + 2 = ( x + b ) ( x + c )Find the value of | b − c |

Question

Given that   a  ,  b  ,  c  ∈      Z    a  &amp;gt;  10and  (  x  −  a  )  (  x  −  12  )  +  2  =  (  x  +  b  )  (  x  +  c  )Find the value of       |    b    −    c    |

bosco84jsrztu · Accepted Answer

Step 1Expanding the quadratic, we have       x    2    −  (  12  +  a  )  x  +  2  =      x    2    +  (  b  +  c  )  x  +  b  c .Setting coefficients equal, we have   −  12  −  a  =  b  +  c and   12  a  +  2  =  b  c .Solving for a in both equations,   a  =  −  12  −  b  −  c  =            b      c      −      2        12   . Multiplying by 12, we have   b  c  +  12  b  +  12  c  +  144  =  (  b  +  12  )  (  c  +  12  )  =  2 .The constraint   a  &amp;gt;  10 gives   b  c  &amp;gt;  122 , and   b  ,  c  &amp;lt;  0 .If b and c are integers, then we must have   b  +  12 and   c  +  12 as   ±  1 and   ±  2 . Therefore b and c are either -10 and -11, or -13 and -14. The first option violates the   b  c  &amp;gt;  122 constraint, but it doesn&#039;t matter, because in either case       |    b  −  c      |    =  1 . The constraint   a  &amp;gt;  10 is actually unnecessary for this problem.

or5a2dosz80z · Accepted Answer

Step 1We need the difference between the roots of the quadratic polynomial   (  x  −  b  )  (  x  −  c  ) .If the quadratic polynomial is   A      x    2    +  B  x  +  C  =  0 then the difference between its roots is                     B        2            −      4      A      C        A   . This is because its roots are             −      B      ±                        B          −                4        A        C                    2      A       .In our case the left-hand side is the polynomial       x    2    −  (  a  +  12  )  x  +  12  a  +  2 .So the difference between its roots, which are b and c, is       (    a    +    12          )      2        −    4    (    12    a    +    2    )   .I will assume the I in your question means integers.Notice that the polynomial inside the square root is       a    2    −  24  a  +  136  =  (  a  −  12      )    2    −  8 . We need this to be a square t. So   8  =  (  a  −  12      )    2    −      t    2    =  (  a  +  t  −  12  )  (  a  −  t  −  12  ) .Now, we consider the factorizations of 8.   ±  1  ×  ±  8   ±  2  ×  ±  4 . We give these values to   a  +  t  −  12 and   a  −  t  −  12 and solve for t and a.The option   ±  1  ×  ±  8 cannot be because we get fractional a. So, we consider the second two. We get   t  =  1 and   a  =  (  6  +  24  )      /    2 .So,       |    b  −  c      |    =  t  =  1 .

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