(16x-6)/(4x^2-3x-1)^0.5

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Solution for (16x-6)/(4x^2-3x-1)^0.5 equation:


D( x )

(4*x^2-(3*x)-1)^0.5 = 0

4*x^2-(3*x)-1 < 0

(4*x^2-(3*x)-1)^0.5 = 0

(4*x^2-(3*x)-1)^0.5 = 0

(4*x^2-3*x-1)^0.5 = 0

4*x^2-3*x-1 = 0

4*x^2-3*x-1 = 0

DELTA = (-3)^2-(-1*4*4)

DELTA = 25

DELTA > 0

x = (25^(1/2)+3)/(2*4) or x = (3-25^(1/2))/(2*4)

x = 1 or x = -1/4

(x+1/4)*(x-1) = 0

(x+1/4)^0.5*(x-1)^0.5 = 0

( x+1/4 )

x+1/4 = 0 // - 1/4

x = -1/4

( x-1 )

x-1 = 0 // + 1

x = 1

4*x^2-(3*x)-1 < 0

4*x^2-(3*x)-1 < 0

4*x^2-3*x-1 < 0

4*x^2-3*x-1 < 0

DELTA = (-3)^2-(-1*4*4)

DELTA = 25

DELTA > 0

x = (25^(1/2)+3)/(2*4) or x = (3-25^(1/2))/(2*4)

x = 1 or x = -1/4

a = 4

a > 0

x in (-1/4:1)

x in (-oo:-1/4) U (1:+oo)

(16*x-6)/((4*x^2-(3*x)-1)^0.5) = 0

(16*x-6)/((4*x^2-3*x-1)^0.5) = 0

4*x^2-3*x-1 = 0

4*x^2-3*x-1 = 0

DELTA = (-3)^2-(-1*4*4)

DELTA = 25

DELTA > 0

x = (25^(1/2)+3)/(2*4) or x = (3-25^(1/2))/(2*4)

x = 1 or x = -1/4

(x+1/4)*(x-1) = 0

(16*x-6)/((x+1/4)^0.5*(x-1)^0.5) = 0

16*x-6 = 0 // + 6

16*x = 6 // : 16

x = 6/16

x = 3/8

x in { 3/8}

x belongs to the empty set

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