posted by Nina on .
If the sides of a triangle have measurements 3x + 4 , 6x - 1, and 8x + 2, find all possible values of x.
I also need help with this question please
All sides must be greater of zero.
Only side: 6x-1 can be negative.
6x>1 Divide both sides with 6
Thanks anyway but that isn't one of the choices the answer is x> 3/11 i figured it out thanks for trying :)
Anonymus solution not completely.
All sides must be greater of zero:
3x> -4 Divide with 3
6x>1 Divide with 6
8x> -2 Divide with 8
Least of that numbers is -4/3= -1.3333
x> -4/3 is solution
x>3/11 is also > -4/3
All x> -4/3 is choices
The condition that all sides have to be positive, as Anonymous used, is not sufficient.
In any triangle the sum of 2 sides must be greater than the third side, so
3x+4 + 6x-1 > 8x + 2 ----> x > -1
3x+4 + 8x + 2 > 6x-1 ---> x > -7/5
8x + 2 + 6x - 1 > 3x+4 --> x > 3/11
the intersection of all three conditions is
x > 3/11
To show that Anonymous is incorrect, pick a value of x between his/her answer of 1/6 and mine of 3/11
e.g. x = 11/50
3x+4 --> 4.66
6x-1 --> .32
8x+2 --> -.24 , contradiction
I have a value of x > 1/6 which did not work.