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Question

Let x1 and x2 be the roots of the equation 2x2+6x+b=0. If x1x2+x2x1=k and b<0, then the minimum integral value of |k| is

A
3.00
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B
3.0
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C
3
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Solution

2x2+6x+b=0
Discriminant of the equation
Δ=368b>0 (b<0)
So, the given equation has two real and distinct roots.
Sum and product of the roots,
x1+x2=3 and x1x2=b2

Now, k=x1x2+x2x1
=x21+x22x1x2 =(x1+x2)22x1x2x1x2 =(x1+x2)2x1x22
k=18b2

We know that,
k<2|k|>2
Hence, the minimum integral value of |k| is 3.

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