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If the equation `ax^(2) + bx + c = 0, a,b, c, in` R have non -real roots, thenA. `c(a - b +c)gt 0`B. `c (a + b + c) gt 0 `C. `c( 4a - 2b + c) gt0 `D. none of these |
Answer» Correct Answer - 1,2,3 Since the roots if ` ax^(2) + bx + c = 0 ` are nonreal , so `f(x) = ax^(2) + bx + c ` Will have same sign for every value of x. Hence, ` f(0) = c. f(1) = a + b + c, f (-1) = a - b +c` ` f(-2) = 4 a - 2b + c ` `rArr c (a +b + c) gt 0, c(a - b + c) gt 0 , c (4a - 2b + c ) gt 0 ` . |
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