N1=8+2*N2
N1*N2=8+2(N1+N2)
one number is 8 more than twice another. their product is 8 more than twice their sum. find the numbers
3 answers
Let the smaller number be x; that means the other number must be 2x + 8.
Solve the equation
x(2x+8) = 2x + 8 + x + 8
which can also be written
2x^2 + 8x = 3x + 16
or
2x^2 + 5x -16 = 0
There seems to be no integer solution. The smaller number is
x = (-5 + sqrt 153)/4 = 1.84233
and the larger number is 2x + 1 = 11.68466
Their sum is 13.5267
and the product is 21.527
as required.
Solve the equation
x(2x+8) = 2x + 8 + x + 8
which can also be written
2x^2 + 8x = 3x + 16
or
2x^2 + 5x -16 = 0
There seems to be no integer solution. The smaller number is
x = (-5 + sqrt 153)/4 = 1.84233
and the larger number is 2x + 1 = 11.68466
Their sum is 13.5267
and the product is 21.527
as required.
drwls, you made a slight mistake - you forgot a factor of 2 in the RHS of the equation.
If N1 = x and N2 = 2x + 8
then the equation is
x(2x+8) = 8 + 2*(2x + 8 + x)
Which simplifies to
x^2 - x - 12 = 0
which has two integer solutions.
Hope that helps!
If N1 = x and N2 = 2x + 8
then the equation is
x(2x+8) = 8 + 2*(2x + 8 + x)
Which simplifies to
x^2 - x - 12 = 0
which has two integer solutions.
Hope that helps!
4 answers