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Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

$g (x) = a (x^{2} + 1) - x (a^{2} + 1)$

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x = - a

and

x = \frac{1}{- a}

x = a

and

x = \frac{1}{a}

x = 2 a

and

x = \frac{1}{2 a}

x = 3 a

and

x = \frac{1}{3 a}

answer is B.

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Detailed Solution

Given quadratic polynomials

g (x) = a (x^{2} + 1) - x (a^{2} + 1)

We put the value of function equal to zero

g (x) = 0

\Rightarrow a (x^{2} + 1) - x (a^{2} + 1) = 0

\Rightarrow a x^{2} + a - a^{2} x - x = 0

\Rightarrow a x^{2} - a^{2} x - x + a = 0

\Rightarrow ax (x - a) - 1 (x - a) = 0

\Rightarrow (x - a) (ax - 1) = 0

(x - a) = 0

and

(ax - 1) = 0

then

x = a

and

x = 1 / a

Now Using this formula,

Sum of zeros = - \frac{coefficient of x}{coefficient of x^{2}}

Put the value in the formula

a + \frac{1}{a} = - \frac{(- (a^{2} + 1))}{a}

\frac{(a^{2} + 1)}{a} = \frac{(a^{2} + 1)}{a}

We calculate

Product of roots

= \frac{constant}{coefficient of x^{2}}

a \times \frac{1}{a} = \frac{a}{a}

1 = 1

Verified the relation between zeros and their coefficients.
Hence, the zeros of each of

g (x) = a (x^{2} + 1) - x (a^{2} + 1)

x = a

and

x = 1 / a

.
Option 2 is correct.

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