If 'a' and 'b' are the non-zero distinct zeros of x2+ax+b, then the least value of quadratic polynomial x2+ax+b is
In a Quadratic polynomial ax2+bx+c, if α and β are the zeros, then
α+β=−ba
αβ=ca
It is given that, 'a,b' are the zeros of x2+ax+b.
⇒a+b=−a--(1)
Product of the roots (ab)=b
⇒a=1 [∵a,b≠0]
Substitute a=1 in a+b=−a
⇒1+b=−1
⇒b=−1−1
⇒b=−2
Hence, the given polynomial is,
x2+ax+b
=x2+x−2---(2)
Since, In ax2+bx+c if a>0 then there exist a least value. and
Least value =4ac−b24a.
Now, In x2+x−2, Co-efficient of x2=1,>0
So, the least value =4(1)(−2)−(1)24(1)
=−8−14
=−94
Therefore, The least value of x2+x−2 is −94
Alternate method for finding least value:
x2+x−2
=(x2+2.12x+(12)2)−(12)2−2
=(x+12)2−14−2
=(x+12)2−94
Since, (x+12)2≥0 [Square of any number either positive or zero]
⇒ Least value =−94
Hence, Option C is correct.