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Mathematics 24 Online
OpenStudy (thomasfire55):

Solve:

OpenStudy (thomasfire55):

a) {-4, 2} b) {-2, 4} c) {}

OpenStudy (philomath):

The first step is to cross multiply : \[\Large \frac{x+4}{4} =\frac{x}{x-2} \] \[\Large 4(x) = (x+4)(x-2)\]

OpenStudy (thomasfire55):

what do you do next?

OpenStudy (philomath):

Simplify on both sides. So you multiply 4 by x to get 4x and (x+4) by (x-2) Do you know how to do that last part ?

OpenStudy (thomasfire55):

not right off

OpenStudy (acespeedfighter):

Het thomas ;)

OpenStudy (philomath):

I'm sorry I'm back. Basicaly you do what is called FOILing in order to find the appropriate expression. You can practice here: http://www.mesacc.edu/~scotz47781/mat120/notes/polynomials/foil_method/foil_method_practice.html Foil stands for: First Outer Inner Last Basically a systematic way to multiply each other together: \[\Large (x+4)(x-2) = (x*x) + (x)(-2) + (4)(x) + (4)(-2)\]

OpenStudy (slenderman):

and at no point in time were you wrong

OpenStudy (philomath):

Basically you get 4x on the left hand side and the RHS is \[\Large (x+4)(x-2) = x^2-2x-8\]

OpenStudy (acespeedfighter):

1

OpenStudy (thomasfire55):

so is the awnser -4 and 2

OpenStudy (philomath):

\[\Large 4x = x^2-2x-8\] Solving for x gets you \[\Large 4x - \color{blue}{4x} = x^2-2x-8 - \color{blue}{4x}\] \[\Large 0 = x^2-6x-8\] (x+2)(x-4) x= -2 x = 4

OpenStudy (thomasfire55):

thank you

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