What are the roots of the quadratic equation of 3x^-2x+1/3=0
Answers
You should insert proper brackets in question. though I had answered this as I interpreted this question.
Answer:
x = 0
Step-by-step explanation:
Solve for x over the real numbers:
1/3 (3 x)^(1 - 2 x) = 0
1/3 (3 x)^(1 - 2 x) = 3^(-2 x) x^(1 - 2 x):
3^(-2 x) x^(1 - 2 x) = 0
Split 3^(-2 x) x^(1 - 2 x) into separate parts with additional assumptions.
Assume x!=0 and -2 x>=0 from x^(-2 x):
3^(-2 x) = 0 for -2 x>=0 and x!=0
or x = 0 for -2 x>=0 and x!=0
or x^(-2 x) = 0
3^(-2 x) = 0 has no solution since for all z element R, 3^z>0:
x = 0 for -2 x>=0 and x!=0
or x^(-2 x) = 0
x = 0 violates the assumption x!=0, which means it is not a valid solution:
x^(-2 x) = 0
Assuming -2 x>0, raise both sides to the power -1/(2 x):
x = 0 for x!=0 and -2 x>=0
x = 0 violates the assumption x!=0, which means it is not a valid solution:
Answer: (no real solutions)
Step-by-step explanation:
hope it helps
use the formula for finding the roots
