1/x-3 + 1/ x+4 = 9/8 find all real solutions
this is equivalent to 8[(x+4) + (x-3)] = 9(x-3) (x+4) solve this quadratic eq and if you get 3 or -4 as solution then reject them thats it
SO NO REAL SOLUTION?
?? first solve this quadratic and tell me the solution you got?
\[\frac{ 7\pm \sqrt{37535} }{ 18 }\]
at first I got 9x^2-7x-116 = 0 but I couldnt figure out how to factor it so I did the quadratic equation plugged in and got that.
so i believe that your calculation are correct so since none of your x is 3 or -4 you have 2 real solutions to this equation
the funny thing is I type that in this dumb online program and it said it was wrong.. and my school created the program as a review program.
anyway thank you for your help.
if you still have any confusion please ask
Is the problem supposed to read as follows: 1/(x-3) + 1/(x+4) = 9/8
\[\frac{ 1 }{ x-3 }+ \frac{ 1 }{ x+4 }=\frac{ 9 }{ 8 }\]
@P->Q Thanks.
I see two real solutions. Did you find them?
no I did not srry
This --> 1/(x-3) + 1/(x+4) = 9/8 cranks out as -29/9 and 4.
srry I gave up Im too tried of this.
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