Question

wx^2+wl/2-wx simplified to wlx/2-wx^2/2

can someone show me how they got there thanks!

Answer 1

is it :

\[wx^2 - (wl/2)\]?

Answer 2

wx^2 - wl/2 - wx *

Answer 3

is this the question?

Answer 4

+wl/2

Answer 5

lol okay ^^"

Answer 6

is it :

wx^2 + (wl/2-wx) ?

Answer 7

yes

Answer 8

are you sure of your answer? because I got in the end :\[wx^2 + wl\]

Answer 9

could you show me how you got wx^2 + wl

Answer 10

steps, first same denominator:\[= (wx^2(2-wx) + wl) /2-wx\]\[= (2wx^2 -w^2x^3 + wl)/2-wx\] then multiply by 2+wx up and down:\[= (2wx^2-w^2x^3+wl)(2+wx)/(2-wx)(2+wx)\]\[= (4wx^2-w^3x^4 + 2wl + w^2xl)/(2-wx)(2+wx)\] then factor by grouping :\[= [wx^2(4-w^2x^2) + wl(2 + wx)]/(2-wx)(2+wx)\]\[= [wx^2(2-wx)(2+wx) + wl(2+wx)]/(2-wx)(2+wx)\]\[= [(2+wx)(2-wx)(wx^2+wl)]/(2-wx)(2+wx)\]

simplify (cancel the commons) :

\[= wx^2 + wl\]

^_^ correct me if I'm wrong please .

Answer 11

thank you!!!

Answer 12

a pleasure :)

Answer 13

do you use a program to type the formulas?

Answer 14

no, click on equation on your reply box and type the formula there ^_^

Answer 15

"equation" *