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
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calculate the values for a ?
- Thread starter raven2k7
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HallsofIvy
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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?
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?
HallsofIvy
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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?)
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?)