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Mathematics 10 Online
OpenStudy (anonymous):

solve sqrt(7x+46)=x+4

OpenStudy (saifoo.khan):

7x + 46 = (x+4)^2

OpenStudy (saifoo.khan):

-6,5

OpenStudy (zarkon):

x=5 is the solution

OpenStudy (saifoo.khan):

are u here sat?

OpenStudy (anonymous):

as we know, you have to check your answer. if you replace x by -6 you get \[\sqrt{4}=-2\] which is false, but if you square both sides you get \[4=4\] which is true. this shows that if you square both sides you can introduce a solution that is not true for the original equation

OpenStudy (saifoo.khan):

ALRIGHTO!

OpenStudy (saifoo.khan):

thanks.

OpenStudy (anonymous):

yw

OpenStudy (saifoo.khan):

so checking is the most important part in this..

OpenStudy (saifoo.khan):

right?

OpenStudy (anonymous):

yes if you square you have to check, because squaring is not "one to one"

OpenStudy (saifoo.khan):

GREAT.. xD

OpenStudy (saifoo.khan):

i would fan u again for this.

OpenStudy (anonymous):

thanks guys!

OpenStudy (anonymous):

you would not have to check if you were taking the log for example, because \[\log(a)=\log(b)\iff a=b\]

OpenStudy (anonymous):

i have another one about the fower garden posted, if i could get some help :)

OpenStudy (anonymous):

but \[a^2=b^2\] does not imply \[a=b\]

OpenStudy (saifoo.khan):

Cool.

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