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KEAM 2017 Math Paper

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Question : 40 of 120

Marks: +1, -0

If a and b are the non zero distinct roots of

x^2 + ax + b = 0,

then the minimum value of

x^2 + ax + b

is

2/3
9/4
– 9/4
– 2/3
1

Solution:

Given,

x^{2}+a x+b=0

For district now zero roots

D > 0

\Rightarrow \;\; a^{2}-4 b>0

Now,

x^{2}+a x+b

=\left(x+\frac{a}{2}\right)^{2}+\left(b-\frac{a^{2}}{4}\right)

=\left(x+\frac{a}{2}\right)^{2}-\left(a^{2}-\frac{4b}{4}\right)

We know, sum of roots

a+b=-a

\Rightarrow \;\; 2a+b=0

.....(i)

Product of roots

a \times b = b

\Rightarrow \;\; b(a-1)=0

\Rightarrow \;\; a=1, b \neq 0

From Eq. (i).

2a+b=0

2(1)+b=0

b=-2

Now,

\left(x+\frac{a}{2}\right)^{2}-\left(\frac{1^{2}+4 \times 2}{4}\right)

=\left(x+\frac{a}{2}\right)^{2}-\frac{9}{4}

\therefore

Minimum value

=-\frac{9}{4}

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