SOLUTION: Can you PLEASE help me figure these out. I am clueless!Thanks in advance! Solve. 1) Square root of 3x-8 +4=6 2)Find the distance between (1,-3)and(4,3)

Click here to see ALL problems on Radicals

Question 125760: Can you PLEASE help me figure these out. I am clueless!Thanks in advance!
Solve.
1) Square root of 3x-8 +4=6

2)Find the distance between (1,-3)and(4,3)
Found 2 solutions by stanbon, checkley71:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Solve.
1) Square root of (3x-8) +4=6
sqrt(3x-8) = 2
Square both sides to get:
3x-8 = 4
3x = 12
x = 4
--------------------
2)Find the distance between (1,-3)and(4,3)
distance = sqrt[(change in x)^2 + (change in y)^2]
distance = sqrt[(4-1)^2 + (3--3)^2]
distance = sqrt[9+36]
distance = sqrt(45)
distance = 3 sqrt(5)
=======================
Cheers,
Stan H.

Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website!
sqrt(3x+8)+4=6
sqrt(3x+8)=6-4
sqrt(3x+8)=2 now square both sides of the equation
3x+8=2^2
3x+8=4
3x=4-8
3x=-4
x=-4/3 answer.
proof
sqrt(3*-4/3+8)+4=6
sqrt(-4+8)+4=6
sqrt(4)+4=6
2+4=6
6=6
----------------------------
first we find the x & y distances between these two coordinates:
x=4-1=3
y=3+3=6
so we have a right triangle with the 2 sides = to 3 & 6.
using the pythagorean theorum
a^2+b^2=c^2 c being the hypotenuse or distance between these two points.
362+662=c^2
9+36=c^2
45=c^2
c=sqrt45
c=6.7 is the distance.