Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively.

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

= \(x^2 – 4x + 1\)

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