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OpenStudy (acxbox22):
x=7
OpenStudy (anonymous):
1) multiply out the denominators (ie. multiply both sides of the equation by x then 7. Do this and tell me what you get.
OpenStudy (anonymous):
you should get \[x ^{2}-49=0\]
OpenStudy (anonymous):
Did you get this?
OpenStudy (anonymous):
yes
what next
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OpenStudy (anonymous):
Are you familiar with the "difference of two squares" rule?
OpenStudy (anonymous):
yes
OpenStudy (acxbox22):
just add 49 to both sides and take its squareroot
OpenStudy (anonymous):
|dw:1397615403507:dw|
OpenStudy (acxbox22):
much easier
you are doing the same thing but in a longer way
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OpenStudy (anonymous):
|dw:1397615457872:dw|
OpenStudy (anonymous):
So can you see that the equation is \[x ^{2}-7^{2}\] this factorises out into \[(x-7)(x+7)=0 \] ...voila... or you can use acxbox22's method!
OpenStudy (anonymous):
Do you understand?
OpenStudy (anonymous):
yes a little whaat is that method like how do i do it
OpenStudy (anonymous):
So, \[x ^{2}-49=0 ... x ^{2}=47 ....x=+ or - \sqrt{47} ....=+ or - 7\]
You simply take the square root. But the crucial step is to say "plus or minus" the square root. Because it could be either - there are two solutions
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