calculate the values for a ?

[raven2k7]

hi, hope someone can help me. the question is asking me to calculate the values for *a* in order for the equation to be equal is 2. the answers for the equation are (a=3 or a=-2) but i cant quite figure out how they were able to factorize the equation in order to arrive to that conclusion. i will upload the equation and what i have done thanks

 

[HallsofIvy]

The numerator is already "factorized". What you did was multiply, the opposite of "factorizing".

Yes, $\displaystyle \frac{(ax- x)(2+ ax)}{3x^2+ 1}= \frac{(ax)(2)+ (ax)^2- 2x- ax^2}{3x^2+ 1}=\frac{(a^2-a)x^2+ (2a- 2)x}{3x^2+ 1}$.

Now we need to take the limit as x goes to infinity. It is difficult to deal with "infinity" but fairly easy to deal with "zero" and as x goes to infinity 1/x goes to 0. So to deal with limits as x goes to infinity, often the best thing to do is to divide both numerator and denominator by the highest power of x in the problem.

Here that highest power of x is $\displaystyle x^2$. Dividing both numerator and denominator by $\displaystyle x^2$ gives $\displaystyle \frac{(a^2- a)\frac{x^2}{x^2}+ (2a+ 2)\frac{x}{x^2}}{3\frac{x^2}{x^2}+ \frac{1}{x^2}}$$\displaystyle = \frac{(a^2- a)+ (2a+ 2)\frac{1}{x}}{3+ \frac{1}{x}}$.

Now, as x goes to infinity the "$\displaystyle \frac{1}{x}$" terms go to 0 so the limit is $\displaystyle \frac{a^2- a}{3}$. So now the question is "What value of a makes $\displaystyle \frac{a^2- a}{3}= 2$?

Can you answer that?

 

[raven2k7]

ok wow , yea i see it now , (-2)*(-2) =-(-2) = 6 ---> 6/3=2 . and if u use 3 you would get the same answer. wow u made that look easy. the only part that confused me is why u factorized that numerator a second time before dividing it by x^2 .

 

[raven2k7]

ok i did checked it and realized that its not necessary , i did it without factorizing again and got the same answer. thanks so much for ur help

 

[HallsofIvy]

I don't know what you mean. I didn't "factorize" anything. I multiplied $\displaystyle (ax- x)(2+ ax)= (ax)(2)+ (ax)(ax)- (x)(2)- (x)(ax)= 2ax+ a^2x^2- 2x- ax^2= a^2x^2- ax^2+ 2ax- 2x= (a^2- a)x^2+ (2a- 2)x$

Well, I guess I did "factorize" "$\displaystyle a^2x^2- ax^2$" to "$\displaystyle (a^2- a)x^2$" and "$\displaystyle 2ax- 2x$" to "$\displaystyle (2a- 2)x$. That was just taking out the common power of x.

(I keep writing "factorize" in quotes because I would use the simpler "factor". Are you British?)

 

[raven2k7]

no i am not. but i am from a British colony, so thats probably why lol . thanks again. this really helped, even the calculator couldn't resolve this type of equation lol