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Question

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

A
23
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B
94
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C
94
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D
1
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Solution

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,b0]

Substitute a=1 in a+b=a

1+b=1

b=11

b=2

Hence, the given polynomial is,

x2+ax+b

=x2+x2---(2)
Since, In ax2+bx+c if a>0 then there exist a least value. and

Least value =4acb24a.

Now, In x2+x2, Co-efficient of x2=1,>0

So, the least value =4(1)(2)(1)24(1)

=814

=94

Therefore, The least value of x2+x2 is 94

Alternate method for finding least value:
x2+x2

=(x2+2.12x+(12)2)(12)22

=(x+12)2142

=(x+12)294

Since, (x+12)20 [Square of any number either positive or zero]
Least value =94

Hence, Option C is correct.


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