Question : Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively.
(i) \(1\over 4\), -1
(ii) \(\sqrt{2}\), \(1\over 3\)
(iii) 0, \(\sqrt{5}\)
(iv) 1, 1
(v) \(-1\over 4\), \(1\over 4\)
(vi) 4, 1
Solution : Let the polynomial be \(ax^2 + bx + c\) and its zeroes be \(\alpha\) and \(\beta\).
(i) Here, \(\alpha\) + \(\beta\) = \(1\over 4\) and \(\alpha\).\(\beta\) = -1
Thus, the polynomial formed = \(x^2\) – (sum of zeroes) x + Product of zeroes
= \(x^2\) – (\(1\over 4\))x – 1
= \(4x^2 – x – 4\)
(ii) Here, \(\alpha\) + \(\beta\) = \(\sqrt{2}\) and \(\alpha\).\(\beta\) = \(1\over 3\)
Thus, the polynomial formed = \(x^2\) – (sum of zeroes) x + Product of zeroes
= \(x^2\) – (\(\sqrt{2}\))x + \(1\over 3\)
= \(3x^2 – 3\sqrt{2}x + 1\)
(iii) Here, \(\alpha\) + \(\beta\) = 0 and \(\alpha\).\(\beta\) = \(\sqrt{5}\)
Thus, the polynomial formed = \(x^2\) – (sum of zeroes) x + Product of zeroes
= \(x^2\) – (0)x + \(\sqrt{5}\)
= \(x^2 + \sqrt{5}\)
(iv) Here, \(\alpha\) + \(\beta\) = 1 and \(\alpha\).\(\beta\) = 1
Thus, the polynomial formed = \(x^2\) – (sum of zeroes) x + Product of zeroes
= \(x^2\) – (1)x + 1
= \(x^2 – x + 1\)
(v) Here, \(\alpha\) + \(\beta\) = \(-1\over 4\) and \(\alpha\).\(\beta\) = \(1\over 4\)
Thus, the polynomial formed = \(x^2\) – (sum of zeroes) x + Product of zeroes
= \(x^2\) – (\(-1\over 4\))x + \(1\over 4\)
= \(4x^2 + x + 1\)
(vi) Here, \(\alpha\) + \(\beta\) = 4 and \(\alpha\).\(\beta\) = 1
Thus, the polynomial formed = \(x^2\) – (sum of zeroes) x + Product of zeroes
= \(x^2\) – (4)x + 1
= \(x^2 – 4x + 1\)
