n^(1/2)+8=n+1

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Solution for n^(1/2)+8=n+1 equation: 


D( n )

n < 0

n < 0

n in <0:+oo)

n^(1/2)+8 = n+1 // - n+1

n^(1/2)-n-1+8 = 0

n^(1/2)-n+7 = 0

t_1 = n^(1/2)

1*t_1^1-1*t_1^2+7 = 0

t_1-t_1^2+7 = 0

DELTA = 1^2-(-1*4*7)

DELTA = 29

DELTA > 0

t_1 = (29^(1/2)-1)/(-1*2) or t_1 = (-29^(1/2)-1)/(-1*2)

t_1 = (29^(1/2)-1)/(-2) or t_1 = (29^(1/2)+1)/2

t_1 = (29^(1/2)-1)/(-2)

n^(1/2)-((29^(1/2)-1)/(-2)) = 0

1*n^(1/2) = (29^(1/2)-1)/(-2) // : 1

n^(1/2) = (29^(1/2)-1)/(-2)

( (29^(1/2)-1)/(-2) < 0 i 1/2 in (0:1) ) => n naleu017Cy do O

t_1 = (29^(1/2)+1)/2

n^(1/2)-((29^(1/2)+1)/2) = 0

1*n^(1/2) = (29^(1/2)+1)/2 // : 1

n^(1/2) = (29^(1/2)+1)/2

n^(1/2) = (29^(1/2)+1)/2 // ^ 2

n = ((29^(1/2)+1)^2)/4

n = ((29^(1/2)+1)^2)/4

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