b के किस मान के लिए समीकरणों `x^(2) + bx -1 = 0, x^(2) + x -b = 0` का एक मूल उभयनिष्ठ होगा?

Explanation

दिए समीकरण, `x^(2) + bx -1 =0`
तथा `x^(2) + x + b -0`
उभयनिष्ठ के लिए,
`(x^(2))/(b xx b + 1 xx 1) = (x)/(1 xx -1 -1 xx b) = (1)/(1 xx 1-1 xx b)`
`(x^(2))/(b^(2) + 1) = (x)/(-1-b) = (1)/(1-b)`
`:. (x^(2))/(b^(2) +1) = (x)/(-1-b)`
`x = (b^(2) + 1)/(-(b +1)) = (-(b +1))/(1-b)`
`(b^(2) +1)(1 -b) = (b + 1)^(2)`
`rArr b^(2) +1 - b^(3) -b = b^(2) + 1 + 2b`
`b^(3) + 3b = 0`
`b(b^(2) + 3) = 0`
`b = 0` या `b^(2) + 3 = 0`
`b^(2) = -3`
`b = +-sqrt-3`
`b = +- sqrt3i`
अत: b `=0, +- sqrt3i`