49^4-k=(1/7)^5-k

Solution for 49^4-k=(1/7)^5-k equation:

49^4-k=(1/7)^5-k

We move all terms to the left:

49^4-k-((1/7)^5-k)=0


Domain of the equation: 7)^5-k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

-k-((+1/7)^5-k)+49^4=0

We add all the numbers together, and all the variables

-1k-((+1/7)^5-k)+5764801=0

We multiply all the terms by the denominator


-1k*7)^5-k)-((+5764801*7)^5-k)+1=0

We add all the numbers together, and all the variables

-1k*7)^5-k)-(40353607^5-k)+1=0

Wy multiply elements

-7k^2=0

a = -7; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-7)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$k=\frac{-b}{2a}=\frac{0}{-14}=0$