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