Thread: modulus

1. simplify (9mod2x-1)-(2mod6x-3) and solve the equation
9mod2x-1 = (9+2mod6x-3)

2.find the values of a and b for which 2x^2+4x-3 can be expressed in the form (2x+a)^2 +b

Originally Posted by amath9

1. simplify (9mod2x-1)-(2mod6x-3) and solve the equation
9mod2x-1 = (9+2mod6x-3)

More parentheses! Do you mean '9 (mod 2) x- 1' or '9 (mod 2x) -1' or '9 (mod 2x-1)'?

Assuming you mean 9 (mod 2x-1)- 2 (mod 6x-3), then saying 9 (mod 2x-1)= y is the same as saying 9= y+ k(2x-1) for some integer k. Multiplying both sides by 3, 27= 3y+ 3k(2x-1)= 3y+ k(6x-3). That is, 27 (mod 6x-1)= 3y so y= 9 (mod 6x-1). 9(mod 2x-1)= 9 (mod 6x-3). 9 (mod 2x-1)- 2 (mod 6x-3)= 9 (mod 6x-3)- 2 (mod 6x-3)= (9- 2) (mod 6x-3)= 7 (mod 6x-3).
The equation 9 (mod 2x-1)= 9+ 2 (mod 6x-3) is the same 9 (mod 2x-1)- 2 (mod 6x-3)= 7 (mod 6x-3)= 9 (mod 6x- 3). That says that, for some integers k, j, n, 7= n+ k(6x-3) and 9= n + j(6x-3). Then n= 7- k(6x-3)= 9- j(6x-3). (j- k)(6x-3)= 9- 7= 2. Since j and k can be any integers, j- k is just some integer. Setting j- k= m, m(6x- 3)= 2. 2 is prime. It only factors are 1 and 2. So either m= 1 and 6x- 3= 2 or m= 2 and 6x-3= 1. If 6x-3= 2, then 6x= 5 and x= 5/6. If 6x-3= 1, then 6x= 4 and x= 4/6= 2/3. x must be either 5/6 or 2/3.

2.find the values of a and b for which 2x^2+4x-3 can be expressed in the form (2x+a)^2 +b

There is no such a and b. No matter what the a and b are, will have leading term and there no such term in . We could express it in the form Set those equal: . Multiply that square on the right: . Now in order that this be true for all x, corresponding coefficients must be equal: we must have 2= 4c, 8= 4ac and -6= . Solve those three equations for a, b, and c.

Hello,

I'm wondering if it's mod as absolute values.. ? It would be more appropriate with the level of the second question ?
But even in this case, we shall need more parentheses...