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Gislervron2

Gislervron2

Answered question

2022-06-02

Given that a , b , c Z
a > 10
and
( x a ) ( x 12 ) + 2 = ( x + b ) ( x + c )
Find the value of | b c |

Answer & Explanation

bosco84jsrztu

bosco84jsrztu

Beginner2022-06-03Added 3 answers

Step 1
Expanding 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 > 10 gives b c > 122 , and b , c < 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 > 122 constraint, but it doesn't matter, because in either case | b c | = 1 . The constraint a > 10 is actually unnecessary for this problem.
or5a2dosz80z

or5a2dosz80z

Beginner2022-06-04Added 1 answers

Step 1
We 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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