2006-07-25 18:03:58 · 6 answers · asked by y0marc 1 in Education & Reference ➔ Higher Education (University +)
Transfer the 4 and square both sides:
x+1 = (2x-4)^2 = 4x*x - 16x + 16
4x^2 -17x +15 =0
and solve as a regular quadratic equation. note that x has to be greater then 1, otherwise the sqrt is invalid
2006-07-25 18:09:22 · answer #1 · answered by mityaj 3 · 0⤊ 0⤋
You use order of operations to seperate the variables,
so first (x+1) = (2x-4)^2, put it in the form of ax^2+bx+c=0 and solve using the quadratic formula x = -b+-SQRT(b^2-4ac)/(2a)
2006-07-26 01:08:33 · answer #2 · answered by Evan K 2 · 0⤊ 0⤋
sqrt(x + 1) + 4 = 2x
sqrt(x + 1) + 4+2 = 2x+2
sqrt(x + 1) + 6 = 2(x +1) suppose sqrt(x+1)=y
y+6= 2y*y
2y*y-y-6=0
(y-2)(2y+3)=0
y=2 so sqrt(x+1)=2
then, x+1=4
so,x=3 the guy under me have a second answer 1.25,but i'm afraid it's wrong.you see, if x=1.25,sqrt(x+1)=-1.5<0 !!!!
wrong,isn't it?
2006-07-26 01:17:16 · answer #3 · answered by photosynthesis 1 · 0⤊ 0⤋
here are the steps:
sqrt(x+1)=2x-4
x+1=(2x-4)^2
x+1=4x^2-16x+16
4x^2-17x+15=0
then by using the quadratic equation:
x will have 2 values:
x=3
x=10/8 (or 1.25)
2006-07-26 01:22:43 · answer #4 · answered by alandicho 5 · 0⤊ 0⤋
parenthesis mean nothing in this situation ignore them
x+1+4=2x
x+5=2x
/x /x
5=x
2006-07-26 01:15:28 · answer #5 · answered by enviroman2222 3 · 0⤊ 0⤋
do your own homework.
It does you no good by leeching answers from here.
2006-07-26 01:06:14 · answer #6 · answered by Iomegan 4 · 0⤊ 0⤋