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Greetings, i need help with this math problem. Thank you

Greetings, i need help with this math problem. Thank you-example-1
User Julienne
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1 Answer

5 votes

The numerator of the left hand side can be rewritten as:


x^2+6x+9=(x+3)^2

Then, the given equation can be written as:


((x+3)^2)/(x+3)=0

Since


(x+3)^2=(x+3)(x+3)

we have


((x+3)(x+3))/(x+3)=0

We can to can cancel out one term x+3 and get


x+3=0

which gives


x=-3

Finally, in order to check that this value corresponds to a real answer, we need to subsitute this value into the equation and compute the limit when x approaches to -3, that is,


lim_(x\rightarrow-3)(x^2+6x+9)/(x+3)

which gives


l\imaginaryI m_{x\operatorname{\rightarrow}-3}(x^(2)+6x+9)/(x+3)=(0)/(0)

Since the limit has the form 0/0 we can to apply L'Hopital rule, that is,


l\imaginaryI m_{x\operatorname{\rightarrow}-3}((d)/(dx)(x^2+6x+9))/((d)/(dx)(x+3))

which gives


l\imaginaryI m_{x\operatorname{\rightarrow}-3}((d)/(dx)(x^(2)+6x+9))/((d)/(dx)(x+3))=l\imaginaryI m_{x\operatorname{\rightarrow}-3}(2x+6)/(1)=(0)/(1)=0

Since the limit exists and is equal to zero then the solution of the equation is: x= -3

User Skyguard
by
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