Thread: Quadratic equations

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!

Originally Posted by spadda

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!

in the first question... what do you mean.. 3^(2x)-4^x+3=0 or (3^2)x-4^x+3=0. either way i can't see how this can be solved using roots of a quadratic.
for the second one please post the solution of yours. if there's is some mistake i will point it out.

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 ...

Originally Posted by spadda

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)

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.

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!

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.

Thanks Halls, I thought the first was a quadractic in 3^x but I couldn't think how to write 4^x in terms of it. Well now I know you can't

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

Originally Posted by spadda

can you please show me this step by step

You have been told what to do. Please show some work and say where you are stuck.

Originally Posted by spadda

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 you committed a mistake here. it should be 3-3(x^2)=8x

3 = 11x

x = 3/11 ...

...