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Maskotti
Sep21-04, 04:27 AM
3.4 10


2x+3\sqrt{3x-5}=5



3\sqrt{3x-5}=5-2x \ \ \ \ \ \ \mid ()^2



\\9(3x-5)=25+2*5*(-2x)+(2x)^2



\\27x-45=25-20x+4x^2



\\27x+20x-4x^2=25+45



\\47x-4x^2=70



\\-4x^2+47x-70=0


hmhph


x= \frac { -47^+_- \sqrt {47^2-4*(-4)*-(70)}}{2*-4}




x= \frac { -47^+_-33}{-8}



x= \ 10 \ or \ x= \ \frac {7}{4}

recon
Sep21-04, 06:26 AM
It seems right.

What exactly is your question here?

HallsofIvy
Sep21-04, 07:30 AM
I hope you realize that writing
\\x \left( x - \frac {47}{4} \right) = \frac{70}{4}
doesn't help you at all: if ab=0 then either a=0 or b= 0 but that only works for "= 0".

You could have written 4x^2- 47x-70=0 and perhaps have reconized that this can be factored: (x-10)(4x-7) but I will confess that I got that by looking at your solution!