
Find the quadratic polynomial whose zeroes are reciprocal of the zeroes of the polynomial $f(x) = a{x^2} + bx + c$ , where $a$ is not equal to zero, $c$ is not equal to zero. Then, find the polynomial?
Answer
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Hint: In this question, we need to find a quadratic polynomial whose zeroes are reciprocal of the zeroes of the given polynomial.
Let the zeroes of the polynomial $f(x) = a{x^2} + bx + c$ be $p$ and $q$ .
And the zeroes of the required polynomial are $P$ and $Q$ .
Then, sum of zeroes of a polynomial $ = \dfrac{{ - b}}{a}$ .
And product of zeroes of a polynomial $ = \dfrac{c}{a}$ .
Complete step-by-step answer:
Given polynomial $f(x) = a{x^2} + bx + c$ , whose zeroes are $p$ and $q$ .
Then, we know, sum of zeroes is given by $p + q = \dfrac{{ - b}}{a}$ and product of zeroes is given by $pq = \dfrac{c}{a}$ .
Now, since, the zeroes of the required polynomial are $P$ and $Q$ and are reciprocal of $p$ and $q$ .
So, they will be of the form $P = \dfrac{1}{p}$ and $Q = \dfrac{1}{q}$ .
Then, the sum of zeroes is given by $ = P + Q = \dfrac{1}{p} + \dfrac{1}{q}$ and the product of zeroes is given by $PQ = \dfrac{1}{{pq}}$ ,
Now, on taking least common multiple and solving, we get, $P + Q = \dfrac{1}{p} + \dfrac{1}{q} = \dfrac{{q + p}}{{pq}}$ .
Putting the values of sum and product of $p$ and $q$ , we get,
Sum of roots of required polynomial will be $P + Q = \dfrac{{\dfrac{{ - b}}{a}}}{{\dfrac{c}{a}}}$ ,
i.e., $P + Q = \dfrac{{ - b \times a}}{{c \times a}}$ ,
we get, $P + Q = \dfrac{{ - b}}{c}$ .
Similarly, product of roots of required polynomial will be $PQ = \dfrac{1}{{pq}} = \dfrac{1}{{\dfrac{c}{a}}}$ .
So, $PQ = \dfrac{a}{c}$ .
Hence, the polynomial becomes ${x^2} + \dfrac{b}{c}x + \dfrac{a}{c} = 0$ or it can be also be written as $c{x^2} + bx + a = 0$ .
Thus, the required polynomial is $g(x) = c{x^2} + bx + a$ .
Note: In any quadratic polynomial, i.e., a polynomial in which coefficient of ${x^2}$ is not equal to zero, sum of its roots $ = \dfrac{{ - b}}{a}$ and product of its roots $ = \dfrac{c}{a}$ .
Roots of a quadratic polynomial are always two, since its degree is $2$ .
Roots can be either real and equal, real and distinct or both complex roots.
Let the zeroes of the polynomial $f(x) = a{x^2} + bx + c$ be $p$ and $q$ .
And the zeroes of the required polynomial are $P$ and $Q$ .
Then, sum of zeroes of a polynomial $ = \dfrac{{ - b}}{a}$ .
And product of zeroes of a polynomial $ = \dfrac{c}{a}$ .
Complete step-by-step answer:
Given polynomial $f(x) = a{x^2} + bx + c$ , whose zeroes are $p$ and $q$ .
Then, we know, sum of zeroes is given by $p + q = \dfrac{{ - b}}{a}$ and product of zeroes is given by $pq = \dfrac{c}{a}$ .
Now, since, the zeroes of the required polynomial are $P$ and $Q$ and are reciprocal of $p$ and $q$ .
So, they will be of the form $P = \dfrac{1}{p}$ and $Q = \dfrac{1}{q}$ .
Then, the sum of zeroes is given by $ = P + Q = \dfrac{1}{p} + \dfrac{1}{q}$ and the product of zeroes is given by $PQ = \dfrac{1}{{pq}}$ ,
Now, on taking least common multiple and solving, we get, $P + Q = \dfrac{1}{p} + \dfrac{1}{q} = \dfrac{{q + p}}{{pq}}$ .
Putting the values of sum and product of $p$ and $q$ , we get,
Sum of roots of required polynomial will be $P + Q = \dfrac{{\dfrac{{ - b}}{a}}}{{\dfrac{c}{a}}}$ ,
i.e., $P + Q = \dfrac{{ - b \times a}}{{c \times a}}$ ,
we get, $P + Q = \dfrac{{ - b}}{c}$ .
Similarly, product of roots of required polynomial will be $PQ = \dfrac{1}{{pq}} = \dfrac{1}{{\dfrac{c}{a}}}$ .
So, $PQ = \dfrac{a}{c}$ .
Hence, the polynomial becomes ${x^2} + \dfrac{b}{c}x + \dfrac{a}{c} = 0$ or it can be also be written as $c{x^2} + bx + a = 0$ .
Thus, the required polynomial is $g(x) = c{x^2} + bx + a$ .
Note: In any quadratic polynomial, i.e., a polynomial in which coefficient of ${x^2}$ is not equal to zero, sum of its roots $ = \dfrac{{ - b}}{a}$ and product of its roots $ = \dfrac{c}{a}$ .
Roots of a quadratic polynomial are always two, since its degree is $2$ .
Roots can be either real and equal, real and distinct or both complex roots.
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