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ihatealgebrasomuch
i have to solve this by factoring... 49=9x^2 ... can i put a sqr. root sign over both sides, get 7=9x, and then get x=7/9???
Well, that's not actually factoring, and you lose the answer x=-7/9
since it says solve by factoring i would say do nt take square root write as 9x^2-49=0 or can be written as \[\Large \implies (3x)^2-(7)^2=0\] now use the formula \[\Large a^2-b^2=(a+b)(a-b)\] in the above we have a=3x ,b=7 just put that in the given formula does this help you ?
yes you can square both sides 49=9x^2 \[\sqrt{49}=\sqrt{9x ^2{}}\] 7=3x divide by 3 so x=3/7 but its plus or minus
however is does say fator 49=9x^2 bring 49 to other side of equal sgn 9x^2-49=0 (3x-7)(3x+7)=0 solve for x 3x-7=0 3x=7 divide by 3 x=7/3 and 3x+7=0 3x=-7 x=-7/3