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daser333 [38]
4 years ago
11

What are the restrictions for a? 2a^2+a-15/5a^2+16a+3

Mathematics
1 answer:
koban [17]4 years ago
8 0

ANSWER


The restrictions are

a\ne -3,a\ne -\frac{1}{5}


EXPLANATION


We were given the rational function,


\frac{2a^2+a-15}{5a^2+16a+3}


The function is defined for all values of a, except



5a^2+16a+3=0


This has become a quadratic trinomial, so we need to split the middle term.


We do that by multiplying the coefficient of x^2 which is 5 by the constant term which is 3. This gives us 15.


The factors of 15 that adds up to 16 are 1 and 15.


We use these factors to split the middle term.




5a^2+15a+a+3=0


We now factor to get,


5a(a+3)+1(a+3)=0


We factor further to get,


(a+3)(5a+1)=0



This implies that,


(a+3)=0,(5a+1)=0


This gives


a=-3,a=-\frac{1}{5}


These are the restrictions.





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