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Mathematics 14 Online
OpenStudy (anonymous):

Which is the extraneous solution of

OpenStudy (anonymous):

OpenStudy (anonymous):

–5 –3 3 5

OpenStudy (anonymous):

x^2-2x-15=0 x^2-5x+3x-15=0 x(x-5)+3(x-5)=0 (x+3)(x-5)=0 x=-3,5 what do you mean by "Extraneous Solutions"

OpenStudy (anonymous):

i dont know thats what the question asked. :/

OpenStudy (anonymous):

There is only one solution for that equation.

OpenStudy (anonymous):

You have to discard the x=5.

OpenStudy (anonymous):

so the answer is -3?

OpenStudy (anonymous):

You can plug the -3 value into the equation and you'll check it by yourself

OpenStudy (anonymous):

how do i do that? :/

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