Question

Find quadratic polynomial whose zero are - q and 1/q

Answer 1

Step-by-step explanation:

It must be q instead of -q..

Sum of zeroes=q+1/q

=q^2+1/q

Product of zeroes=q*1/q

=1

The quadratic polynomial will be as follows:

=x^2-(Sum of zeroes)x+(Product of zeroes)

=x^2-(q^2+1/q)x+1

We will take LCM

=qx^2-xq^2-x+1

Answer 2

Answer:

The quadratic polynomial is $qx^2+(1-q^2) x-q =0$ whose zeros are $-q$ and $1/q$.

Step-by-step explanation:

Consider the general form of quadratic polynomial as follows:

$ax^2+bx+c=0$ . . . . . (1)

Let $\alpha$ and $\beta$ be the $zeros$ of the polynomial (1).

Then, the quadratic polynomial (1) can also be written in the form as

$x^2+(\alpha +\beta )x+\alpha \beta =0$ . . . . . (2)

where $\alpha +\beta$ is the sum of zeros and $\alpha \beta$ is the product of zeros.

Consider the given zeros as follows:

$\alpha =-q$  and $\beta =\frac{1}{q}$

Then, sum of zeros,

⇒ $\alpha +\beta = -q +\frac{1}{q}$

             $=\frac{-q^2+1}{q}$

             $=\frac{1-q^2}{q}$

Product of roots,

⇒ $\alpha \beta = (-q)\frac{1}{q}$

         $=-1$

Substitute the values $\frac{1-q^2}{q}$ for $(\alpha +\beta )$ and $-1$ for $\alpha \beta$ in the equation (2) as follows:

$x^2+(\frac{1-q^2}{q} )x+(-1) =0$

⇒ $x^2+\frac{1-q^2}{q} x-1 =0$

Further, simplify as follows:

$qx^2+(1-q^2) x-q =0$

Therefore, the quadratic polynomial is $qx^2+(1-q^2) x-q =0$ whose zeros are $-q$ and $1/q$.

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