Question

find the real and imaginary solutions to the following equation:
7(l4x-1l)-10=(l4x-1l)^2... the lines in the parenthesis are absolute value signs

Answer 1

Let W=4x-11

\(\large \rightarrow 7|W|-10=|W|^2\).

It doesn't matter for the W^2 if W is positive or negative, so there are only two possibilities:

\(\large 7W-10=W^2\), or \(\large -7W-10=W^2\).

Solve each of those for W, then substitute 4x-11 back in to find x.

Answer 2

i am confused on how to solve for w here

Answer 3

They are quadratic equations. You can put them in the form aW^2+bW+c=0, then factor or use quadratic formula.

Answer 4

can you please start me off im am new with this and still confused. thanks

Answer 5

Starting with \(7W−10=W^2\), rearrange it to be \(W^2-7W+10=0\).
This can be factored to \((W-2)(W-5)=0\).

You can the sub back in the W=4x-11 to get 4x-11=2 and 4x-11=5.

Answer 6

but when you bring w^2 to the other side wouldnt it be negative

Answer 7

I didn't bring W^2 to the other side, I brought everything else over to be with W^2.

Answer 8

nevermind i understood that part

Answer 9

but it is not 4x-11... it is 4x-1

Answer 10

LOL, sorry, the || were making my eyes cross.

Answer 11

yea sorry about that

Answer 12

It still works the same way. I'm redoing it to get the correct answers now.

Answer 13

I found four real solutions, but no imaginary ones.

Answer 14

i only found 2 real solutions... x=3/4 and x=1.5

Answer 15

Ok, those are the correct ones from the first equation, now you need to do the same procedure on \(W^2+7W+10=0\).

Answer 16

what were your final solutions?

Answer 17

x \(\epsilon\) {3/4, 3/2, -1/4, and one other one that I'll let you get on your own}

Answer 18

i do not understand how to get -1/4

Answer 19

\[ x=a+ib \]
\[ 7(|4x-1|)-10=|4x-1|^2 \]
\[ 7\sqrt{(4a-1)^2+(4b)^2}-10=\sqrt{(4a-1)^2+(4b)^2}^2 \]
\[ 7\sqrt{(4a-1)^2+(4b)^2}=16a^2-8a+1+16b^2+10 \]

Answer 20

i need to use u-substitution

Answer 21

CliffSedge may you please help me find the last two real solutions?

Answer 22

Same as before.

\(W^2+7W+10=0=(W+2)(W+5)\)

\(\rightarrow 4x-1=-2\), and \(4x-1=-5\)

I'm going through henpen's equation to see if I can find any imaginary solutions.

Answer 23

my final answers came out to be x=3/4, 1.5, -1, -1/4

Answer 24

is this right?

Answer 25

That's what I got too. Well done.

Answer 26

Were you able to make any sense of what henpen posted above?