Hey guys, I've been having trouble with these problems!! iv tried it a million ways and still cant figure it out
any help??
Solve each quadratic equation by factoring:
1. 3^2x - 4^x + 3 = 0
(im completely lost on this one)
2. 1/x - x = 8/3
(for this one i tried multiplying X and 3 to all the parts of the question but i came up with the wrong answer.)
Thanks in advance!
http://desmond.yfrog.com/Himg146/sca...=640&ysize=640
(i print screened the question if it makes any more sense :S)
as for number two, i probably did this totally wrong ...haha
1/x - x = 8/3
1(3) - x(3x) = 8(x)
3 - 3x = 8x
3 = 11x
x = 3/11 ...
Please, please, please use parentheses! You mean 3^(2x)- 4^x+ 3= 0. That is NOT a quadratic and cannot be made into one. But I suspect that you really mean 3^(2x)- 3^x+ 3= 0. If you take y= 3^x, that is the same as y^2- y+ 3= 0. That is a quadratic equation which can be solved by completing the square or the quadratic formula. Once you have found y, solve 3^x= y by taking logarithms of both sides.
If you multiply both sides by 3x you get 3- 3x^2= 8x which is the same as 3x^2+ 8x- 3=0. Again, you can solve that by completing the square or the quadratic formula.2. 1/x - x = 8/3
(for this one i tried multiplying X and 3 to all the parts of the question but i came up with the wrong answer.)
Thanks in advance!
Please, please, please use parentheses! You mean 3^(2x)- 4^x+ 3= 0. That is NOT a quadratic and cannot be made into one. But I suspect that you really mean 3^(2x)- 3^x+ 3= 0. If you take y= 3^x, that is the same as y^2- y+ 3= 0. That is a quadratic equation which can be solved by completing the square or the quadratic formula. Once you have found y, solve 3^x= y by taking logarithms of both side.
can you please show me this step by step