SOLUTION: I have several radical equations I can't figure out how to do. Can I get help on these two: square root 2-x-4=x answer choices: a. -5 b. -2 c. -7 The square root si

Question 91611: I have several radical equations I can't figure out how to do. Can I get help on these two:
square root 2-x-4=x
answer choices:
a. -5
b. -2
c. -7
The square root sign is only over the 2-x part of the equation
The second problem is:
square root 3x^2+4-2=x
answer choices:
a. (0,1) these are the squiggly parenthesis
b. (0,2) these are the squiggly parenthesis
c. 2
The square root sign is only over the 3x^2+4 part of the equation
Thank you for your help
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
square root 2-x-4=x
answer choices:
a. -5
b. -2
c. -7
The square root sign is only over the 2-x part of the equation
------------
sqrt(2-x) - 4 = x
sqrt(2-x) = x+4
Square both sides to get:
2-x = x^2+8x+16
x^2+9x+14=0
(x+2)(x+7)=0
x = -2 or x = -7
---------
Checking out those answers in the original equation:
sqrt(2-x) - 4 = x
If x = -2 you get sqrt(0) -4 = -2 which is false
If x = -7 you get sqrt(9) -4 = -7 which is true
---------
Final solution: x=-7
=============================
The second problem is:
square root 3x^2+4-2=x
sqrt(3x^2+4) - 2 = x
sqrt(3x^2+4) = x+2
Square both sides to get:
3x^2+4 = x^2+4x+4
2x^2-4x=0
2x(x-2) = 0
x=0 or x=2
------------
Checking out those answer in the original equation:
sqrt(3x^2+4) - 2 = x
If x = 0 you get sqrt(4) -2 = 0 which is true
If x = 2 you get sqrt(16) -2 = 2 which is true
--------------
Final solution: x=0 or x=2
==============================
Cheers,
Stan H.
answer choices:
a. (0,1) these are the squiggly parenthesis
b. (0,2) these are the squiggly parenthesis
c. 2
The square root sign is only over the 3x^2+4 part of the equation

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