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If `a and b(!=0)` are the roots of the equation `x^2+ax+b=0` then the least value of `x^2+ax+b` isA. `(9)/(4)`B. `-(9)/(4)`C. `-(1)/(4)`D. `(1)/(4)` |
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Answer» Correct Answer - B Since a and b are the roots of the equation `x^(2) + ax + b = 0`. `therefore" "a+b =-a and ab = b` Now, `ab = b rArr (a-1)b=0 rArr a=1" "[therefore b ne 0]` Putting a = 1 in a + b = - a, we get b = - 2. Clearly, `y=x^(2)+ax + b` is a parabola opening upward. `therefore" "y_(min)=-(D)/(4)" "[therefore y_(min)=-(D)/(4a)"for" y=ax^(2)+bx+c]` `rArr" "y_(min)=-(a^(2)-4b)/(4)=-(9)/(4)` |
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