If `aa n db(!=b)` are the roots of the equation `x^2+a x+b=0,` then find the least value of `x^2+a x+b(x in R)dot`

Options

A. `(9)/(4)`
B. `-(9)/(4)`
C. `-(1)/(4)`
D. `(1)/(4)`

Answer

B

Explanation

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)`