Question

xabs(x-6) = 45x+7. Solve for x.

Answer 1

\[x |x-6|=45x+7\]
right?

Answer 2

yeah

Answer 3

\[|x-6|=\frac{45x+7}{x}\]

Answer 4

is that cool with you I divided by x

Answer 5

y

Answer 6

*yup

Answer 7

now we have two equations to consider

Answer 8

since |x-6|=x-6 if x>6
and |x-6|=-(x-6) if x<6

Answer 9

that is we have the following two equations to solve:
\[x-6=\frac{45x+7}{x} \\ -(x-6)=\frac{45x+7}{x}\]

Answer 10

the main reason I divided both sides by x initially
is that |x-6| can only result in a positive output
so that means (45x+7)/x should be positive

Answer 11

now what I'm going to up doing with these two equations is multiply the x again on both sides

Answer 12

\[x(x-6)=45x+7 \text{ or } -x(x-6)=45x+7 \]

Answer 13

now when we solve both of these x's we should verify them by plugging them into the initial equation

Answer 14

both of these equations are quadratic equations

Answer 15

they may not look it to you right now
but if you use the distributive property a little bit you will see it

Answer 16

yuck these result in some ugly some answers

Answer 17

the two quadratic equations you get are x^2 -51x -7 and -x^2 -39x +7 right?

Answer 18

i meant x^2 -51x -7 and -x^2 -39x -7

Answer 19

yes
\[x(x-6)=45x+7 \\ x^2-6x=45x+7 \\ x^2-6x-45x-7=0 \\ x^2-51x-7=0 \\ \text{ other equation is } -x(x-6)=45x+7 \\ -x^2+6x=45x+7 \\ -x^2+6x-45x-7=0 \\ -x^2-39x-7=0 \\ x^2+39x+7=0\]

Answer 20

so what do you do next?

Answer 21

solve both quadratics

Answer 22

I would probably end up using the approximation for the exact solutions (solutions may be extraneous or not at this point)
that is why I'm using the approximations to find out if it is or isn't...
so after I find all four exacts
I will find approximation to each of those exacts
and then plug them in to the (45x+7)/x thing to see if they are positive or not

Answer 23

it looks like you should wind up with 3 solutions

Answer 24

if i didn't make a mistake with my calculator

Answer 25

You can tell me how far you have gotten
or show me anything you want from this problem and i can tell you if that is what i got

Answer 26

or if you just want to check something

Answer 27

like what approximations for the 4 numbers are you using?

Answer 28

\(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x|x-6| = 45x+7\)

\(x(x-6) = 45x+7~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x[-(x-6)] = 45x+7\)

\(x^2-6x = 45x+7~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-x^2 + 6x = 45x+7\)

\(x^2-51x -7 = 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-x^2 - 39x -7 = 0\)

\(x^2-51x -7 = 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x^2 + 39x +7 = 0\)

\(x = \dfrac{51 \pm \sqrt{(-51)^2 - 4(1)(-7)}}{2(1)} ~~~~~~~~~~~~~
x = \dfrac{-39 \pm \sqrt{(39)^2 - 4(1)(7)}}{2(1)}\)

\(x = \dfrac{51 \pm \sqrt{2629}}{2} ~~~~~~~~~~~~~
x = \dfrac{-39 \pm \sqrt{1493}}{2}\)

Now you need to check each of the four above solutions to see if it works in the original equation.

Answer 29

have to give you points @mathstudent55 for that cute layout