T O P I C R E V I E W 
Nichelle1012 
Posted  09/25/2013 : 20:37:43 In the equation of (2x1)(2x+1)=35, x=3 or 3. Note: The original equation was cross multiplication, which was 2x1 over 7 = 5 over 2x+1. OR 2x1/ 7 = 5/ 2x+1. That simplifies to (2x1)(2x+1)=35.
Now, in another equation, which is a over 16 = 4 over a OR a/ 16 = 4/ a, the answer is a=4.
Why is the first solution x= 3 or 3 (negative or positive) and the second solution is a=4 (only positive)? Why is the answer positive or negative for one equation and only positive for the other? 
3 L A T E S T R E P L I E S (Newest First) 
Nichelle1012 
Posted  09/26/2013 : 17:05:50 quote: Originally posted by the_hill1962
In the first one, you have the quadratic equation 4x1 = 35 that is 4x=36 and finally x=9 which solves to x=+3 and x=3 The reason, 9 is 3 (3*3=9) and 3 (3*3=9)
In the second one, you have a cubic equation a=64 The solution would be a = 64 and that is only +4 because 4*4*4=64. Note that 4*4*4 = 64 and thus, not a cube root of "64"
Basically what is going on is that when you square a number, a negative square is positive so the negative is also an answer. When you cube a number, a negative number cubed is not positive so it is not a solution.
THANKS the_hill1962. You explained everything as if you were a professor. :) 
the_hill1962 
Posted  09/26/2013 : 16:09:54 In the first one, you have the quadratic equation 4x1 = 35 that is 4x=36 and finally x=9 which solves to x=+3 and x=3 The reason, 9 is 3 (3*3=9) and 3 (3*3=9)
In the second one, you have a cubic equation a=64 The solution would be a = 64 and that is only +4 because 4*4*4=64. Note that 4*4*4 = 64 and thus, not a cube root of "64"
Basically what is going on is that when you square a number, a negative squared is positive so the negative number is also an answer since it gives you what you want (a positive). When you cube a number, a negative number cubed is not positive so it is not a solution. 
royhaas 
Posted  09/26/2013 : 11:07:23 In the first equation, a = 3 or a = 3 satisfies the original equation. In the second equation, only a = 3 satisfies the original equation. 

