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If a, b, c are positive and a = 2b + 3c, then roots of the equation `ax^(2) + bx + c = 0` are real forA. `|(a)/(c)-11| ge 4 sqrt(7)`B. `|(c)/(a)-11| ge 4 sqrt(7)`C. `|(b)/(c)+4| ge 2 sqrt(7)`D. `|(c)/(b)-4| ge 2 sqrt(7)` |
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Answer» Correct Answer - A For real roots, we must have `b^(2) - 4ac ge 0` `rArr" "((a-3c)/(2))^(2)-4ac ge 0` `rArr" "a^(2)+9c^(2)-6ac-16ac ge 0` `rArr" "((a)/(c))^(2)-22((a)/(c))+9 ge 0 rArr ((a)/(c)-11)^(2) ge (4 sqrt(7))^(2)` `rArr" "|(a)/(c)-11| ge 4 sqrt(7)` Again, `b^(2) - 4ac ge 0 and a = 2b + 3c` `rArr" "b^(2)-4x(ab+3c)ge0` `rArr" "b^(2)-8bc-12c^(2) ge 0` `rArr" "((b)/(c))^(2)-8((b)/(c))-12 ge 0` `rArr" "((b)/(c)-4)^(2) ge (2 sqrt(7))^(2) rArr |(b)/(c)-4| ge 2 sqrt(7)` Hence, options (a) is true. |
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