Question

Can someone please help me with these last two questions?!

What is the restriction on the quotient of quantity 5 x squared minus 10 x minus 15 divided by quantity x squared minus 9 divided by quantity 6 x plus 12 divided by quantity x plus 2?

Answer 1

@ganeshie8 @Callisto @jhonyy9 @radar

Answer 2

it is good if u will type this equation.

Answer 3

Okay, give me a second.

\[\frac{ 5x^2-10x-15 }{ x^2-9 }\div \frac{ 6x+12 }{ x+12 }\]

Answer 4

for restrictions u set the denominator factors equal to 0.

solve below :
x^2 - 9 = 0
6x + 12 = 0
x + 12 = 0

Answer 5

those give u all the values of x that make the given complex fraction undefined

Answer 6

in other words, those values make the denominator equal to 0

Answer 7

So 3 or -3?

Answer 8

hmmm... ganeshi .. i was just wandering in the question it says "6 x plus 12 divided by quantity x plus 2"
so.. if u take a 6 out from the numerator u can cancel out the numerator and denominator..

also if u factorized the numerator of the first fraction u can cancel out the ( x -3) term from that fraction too.. leaving with u only ( x + 3) as the denominator... then there will be only one restriction..

im saying this cause the question says "What is the restriction on the quotient" not "What are the restrictions on the quotient" so.. i think there should be only one... ignore this if im wrong

thanks in advance :)

Answer 9

@ganeshie8

Answer 10

I don't think that 6x+12=0 needs to be considered, in my humblest opinion.

Answer 11

well then... anything--?

Answer 12

Solve x^2-9=0 and x+12=0, as what ganeshie has said

Answer 13

9.

Answer 14

\[x \neq 9\]

Answer 15

in my point of view ... u should factorize this and simplify it first... it will leave with u only a single term for denominator... let's say its ( x-a) ... then
( x- a) = 0
then the restriction will be
x is not equal to zero

Answer 16

note that that's x^2-9=0 instead of x-9=0

Answer 17

\[x \neq 9\]

Answer 18

I mean -2

Answer 19

kc_kennylau... y cant we simplify this first and go to restrictions afterwards ... can u explain it ?
thanks .

Answer 20

\(\large \frac{ 5x^2-10x-15 }{ x^2-9 }\div \frac{ 6x+12 }{ x+12 } = \frac{\frac{ 5x^2-10x-15 }{ x^2-9 }}{ \frac{ 6x+12 }{ x+12 }}\)

Answer 21

clearly 6x+12 = 0 will make the expression meaningless
so 6x+12 = 0 is also a restriction @kc_kennylau

Answer 22

Oh, I apologize for my mistake.

Answer 23

@***[ISURU]*** we need to check restrictions before simplifying

cancelling of factors just masks the restrictions...

Answer 24

But if so, you can just invert the second fraction, so x+12=0 will not cause any trouble at all.

Answer 25

the only answers I can pick are:

\[x \neq -2 \]
\[x \neq -3 \]
\[x \neq 3 \]
\[x \neq 9 \]

Answer 26

@kc_kennylau