SOLUTION: help?! please show all work so i know how to actually do these. solve for the unknown and check the work. 1. 4√x=20 2. √y+√y+7=7 3. √x+11=√x+1

Algebra ->  Radicals -> SOLUTION: help?! please show all work so i know how to actually do these. solve for the unknown and check the work. 1. 4√x=20 2. √y+√y+7=7 3. √x+11=√x+1       Log On


   

Question 463702: help?! please show all work so i know how to actually do these.
solve for the unknown and check the work.
1. 4√x=20
2. √y+√y+7=7
3. √x+11=√x+1
if its confusing.. in number 2 the √y+7 is a square root
and in number 3 the x+11 is one square root and x+1 is another one. just so you dont think that just the y is or just the x. (:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
1. 4√x=20
2. √y+√y+7=7
3. √x+11=√x+1
if its confusing.. in number 2 the √y+7 is a square root
and in number 3 the x+11 is one square root and x+1 is another one. just so you dont think that just the y is or just the x
...
1. 4√x=20
divide by 4
√x=5
square both sides
x=25
Check:
4√25=4*5=20
..
2. √y+√(y+7)=7
square both sides; ie,(a+b)^2=a^2+2ab+b^2
y+2√y*√(y+7)+y+7=49
y+2√(y(y+7))+y+7=49
2√(y^2+7y)=49-7-2y
2√(y^2+7y)=42-2y
square both sides again
4(y^2+7y)=1764-168y+4y^2
4y^2+28y=1764-168y+4y^2
196y=1764
y=1764/196=9
Check:
√y+√(y+7)=7
√9+√(9+7)=√9+√16=3+4=7
..
3. √(x+11)=√(x+1)
square both sides
x+11=x+1
11≠1
no solution