Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\dfrac{1}{4}, -1\)
(ii) \(\sqrt{2}, \dfrac{1}{3}\)
(iii) \(0, \sqrt{5}\)
(iv) \(1, 1\)
(v) \(-\dfrac{1}{4}, \dfrac{1}{4}\)
(vi) \(4, 1\)
Method using sum and product of zeroes
Let a quadratic polynomial have zeroes \(\alpha\) and \(\beta\). Then
\[ \alpha + \beta = S, \quad \alpha\beta = P. \]
The standard quadratic polynomial whose zeroes are \(\alpha\) and \(\beta\) is
\[ p(x) = x^2 - Sx + P. \]
Any non-zero scalar multiple of this polynomial represents the same pair of zeroes. The given first number is taken as the sum \(S\) and the second as the product \(P\).
(i) \(S = \dfrac{1}{4}, \; P = -1\)
Start with
\[ p(x) = x^2 - \dfrac{1}{4}x - 1. \]
To eliminate fractions, multiply the entire polynomial by 4 (a non-zero constant):
\[ 4p(x) = 4x^2 - x - 4. \]
So a convenient quadratic polynomial is
\[ p(x) = 4x^2 - x - 4. \]
This has sum of zeroes \(\dfrac{1}{4}\) and product \(-1\).
(ii) \(S = \sqrt{2}, \; P = \dfrac{1}{3}\)
Take
\[ p(x) = x^2 - \sqrt{2}x + \dfrac{1}{3}. \]
Multiply through by 3 to clear denominator:
\[ 3p(x) = 3x^2 - 3\sqrt{2}x + 1. \]
Thus one suitable polynomial is
\[ p(x) = 3x^2 - 3\sqrt{2}x + 1. \]
Its sum of zeroes is \(\sqrt{2}\) and product is \(\dfrac{1}{3}\).
(iii) \(S = 0, \; P = \sqrt{5}\)
Using the general form,
\[ p(x) = x^2 - Sx + P = x^2 - 0 \cdot x + \sqrt{5} = x^2 + \sqrt{5}. \]
Here there is no need to multiply by any constant, so
\[ p(x) = x^2 + \sqrt{5} \]
is the required quadratic polynomial whose zeroes have sum 0 and product \(\sqrt{5}\).
(iv) \(S = 1, \; P = 1\)
The polynomial is
\[ p(x) = x^2 - Sx + P = x^2 - x + 1. \]
The sum of its zeroes is 1 and product is 1, as required.
(v) \(S = -\dfrac{1}{4}, \; P = \dfrac{1}{4}\)
First write
\[ p(x) = x^2 - Sx + P = x^2 - \left(-\dfrac{1}{4}\right)x + \dfrac{1}{4} = x^2 + \dfrac{1}{4}x + \dfrac{1}{4}. \]
Multiply by 4 to remove fractions:
\[ 4p(x) = 4x^2 + x + 1. \]
So one convenient polynomial is
\[ p(x) = 4x^2 + x + 1. \]
This polynomial has sum of zeroes \(-\dfrac{1}{4}\) and product \(\dfrac{1}{4}\).
(vi) \(S = 4, \; P = 1\)
From the formula,
\[ p(x) = x^2 - Sx + P = x^2 - 4x + 1. \]
So
\[ p(x) = x^2 - 4x + 1 \]
is a quadratic polynomial whose zeroes have sum 4 and product 1.
Therefore, suitable quadratic polynomials for each pair \((S, P)\) are:
(i) \(4x^2 - x - 4\), (ii) \(3x^2 - 3\sqrt{2}x + 1\), (iii) \(x^2 + \sqrt{5}\), (iv) \(x^2 - x + 1\), (v) \(4x^2 + x + 1\), (vi) \(x^2 - 4x + 1\).