
Steve S. answered 02/28/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
16x - 16 = -1/x
16x^2 - 16x + 1 = 0
a = 16, b = -16, c = 1
x = (-b ± √(b^2 - 4ac))/(2a)
x = (-(-16) ± √((-16)^2 - 4(16)(1)))/(2(16))
x = (16 ± √((16)(16 - 4)))/(2(16))
x = (16 ± √((16)(4)(4 - 1)))/(2(16))
x = (16 ± 8√(3))/(2(16))
x = (2 ± √(3))/4
check:
16((2 ± √(3))/4) - 16 =? -1/((2 ± √(3))/4)
16(((2 ± √(3))/4) - 4/4) =? -4(-2 ± √(3))/((2 ± √(3))(-2 ± √(3)))
4( -2 ± √(3) ) =? -4(-2 ± √(3))/(-4 + 3)
4( -2 ± √(3) ) =? 4(-2 ± √(3)) √