Answer
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Hint- In the quadratic equation, \[a{x^2} + bx + c = 0\]the sum of the roots is given by \[ - \dfrac{b}{a}\]whereas their products is given \[\dfrac{c}{a}\].
In this question, it has already been given that the roots are reciprocal of each other and we need to derive a condition which will satisfy the condition by using the concept of product of the roots.
Complete step by step solution:
Let one of the roots of the quadratic equation $ p(x) = a{x^2} + bx + c $ be $ m $ then, according to the question, the second root of the polynomial will be $ \dfrac{1}{m} $ as the reciprocal of the first one.
The product of the roots of the quadratic equation is the ratio of the coefficient of ${x^0}$ and $ {x^2} $.
Here, the coefficient of $ {x^2} $ is $ a $ while the coefficient of $ {x^0} $ is $ c $.
Hence,
$
m \times \dfrac{1}{m} = \dfrac{c}{a} \\
1 = \dfrac{c}{a} \\
c = a \\
$
Hence, $c = a$ is one of the conditions that the zeroes of the polynomial $ p(x) = a{x^2} + bx + c $ are reciprocal of each other.
Additional Information: Quadratic equation is given as \[a{x^2} + bx + c = 0\] this is the basic equation that contains a squared variable \[x\]and three constants a, b and c. The value of the \[x\] in the equation which makes the equation true is known as the roots of the equation. The numbers of roots in the quadratic equations are two as the highest power on the variable of the equation is x. The roots of the equation are given by the formula\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], where \[{b^2} - 4ac\]tells the nature of the solution.
Note: The equation could have one more condition if we follow the concept of the summation of the roots, given as \[ - \dfrac{b}{a}\] but this will result in another quadratic equation to be solved for the value of the roots.
In this question, it has already been given that the roots are reciprocal of each other and we need to derive a condition which will satisfy the condition by using the concept of product of the roots.
Complete step by step solution:
Let one of the roots of the quadratic equation $ p(x) = a{x^2} + bx + c $ be $ m $ then, according to the question, the second root of the polynomial will be $ \dfrac{1}{m} $ as the reciprocal of the first one.
The product of the roots of the quadratic equation is the ratio of the coefficient of ${x^0}$ and $ {x^2} $.
Here, the coefficient of $ {x^2} $ is $ a $ while the coefficient of $ {x^0} $ is $ c $.
Hence,
$
m \times \dfrac{1}{m} = \dfrac{c}{a} \\
1 = \dfrac{c}{a} \\
c = a \\
$
Hence, $c = a$ is one of the conditions that the zeroes of the polynomial $ p(x) = a{x^2} + bx + c $ are reciprocal of each other.
Additional Information: Quadratic equation is given as \[a{x^2} + bx + c = 0\] this is the basic equation that contains a squared variable \[x\]and three constants a, b and c. The value of the \[x\] in the equation which makes the equation true is known as the roots of the equation. The numbers of roots in the quadratic equations are two as the highest power on the variable of the equation is x. The roots of the equation are given by the formula\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], where \[{b^2} - 4ac\]tells the nature of the solution.
Note: The equation could have one more condition if we follow the concept of the summation of the roots, given as \[ - \dfrac{b}{a}\] but this will result in another quadratic equation to be solved for the value of the roots.
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