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OpenStudy (owlfred):
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OpenStudy (anonymous):
(x+1)(3x-5) = 4(5x-3)
OpenStudy (anonymous):
okay......can you keep going?
OpenStudy (anonymous):
3x^2 -2x -5 = 20x -12
3x^2 - 22x +7 = 0
OpenStudy (anonymous):
does the last line factor into (x-7)(3x-1)?
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OpenStudy (anonymous):
yes
OpenStudy (anonymous):
x = 7 or 1/3
now substitute both these values into ur initial eqn to double check
OpenStudy (anonymous):
okay and on the first line you wrote 4(5x-3).....how did you get the four? 2^2?
OpenStudy (anonymous):
yes 2 can be written as log 4 to base 2
OpenStudy (anonymous):
and then we use log a + log b =log (ab)
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OpenStudy (anonymous):
thanks you. thanks were i kept going wrong. The solutions both work. :)
OpenStudy (anonymous):
no prob miss
OpenStudy (anonymous):
can i ask one more thing? if you are solving for x and you take the square root to isolate x, do you include both the positive and negative answers?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
like lets say x^2 = k
then x = k AND -k
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OpenStudy (anonymous):
in some special situations we can neglect the negative answer..bt generally it is essential to consider both