In quadratic equation `ax^(2)+bx+c=0`, if discriminant `D=b^(2)-4ac`, then roots of quadratic equation are:
A
`1/5`
B
`2/5`
C
`4/5`
D
`3/5`
Text Solution
Verified by Experts
The correct Answer is:
C
For maximum/ minimum value `(dy)/(dx)=0rArr 5(2x)-2(1)+0=0rArr x=1/5` Now at `x=1/5, (d^(2)y)/(dx^(2))=10` which is positive so minima at `x=1/5` Therefore `y_("min")=5(1/5)^(2)-2(1/5)+1=4/5`
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