SOLUTION: Would you be so kind as to help with this. I have to solve these equations: a. (radicand 2x + 1) = -5 b. (radicand x + 1) = x + 1 c. x ^ 1/6

SOLUTION: Would you be so kind as to help with this. I have to solve these equations: a. (radicand 2x + 1) = -5 b. (radicand x + 1) = x + 1 c. x ^ 1/6 - 2 = 0 thank you

Question 546017: Would you be so kind as to help with this.
I have to solve these equations:
a. (radicand 2x + 1) = -5
b. (radicand x + 1) = x + 1
c. x ^ 1/6 - 2 = 0

thank you
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
a. (radicand 2x + 1) = -5
√(2x+1)=-5
square both sides
2x+1=25
2x=24
x=12
..
b. (radicand x + 1) = x + 1
√(x+1)=(x+1)
square both sides
x+1=x^2+2x+1
x^2+x=0
x(x+1)=0
x=0
or
x=-1
..
c. x ^ 1/6 - 2 = 0
x^(1/16)=2
raise both sides to the 16th power
x=2^16)=65536
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