For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.
(i). \(-\dfrac{8}{3}\), \(\dfrac{4}{3}\)
(ii). \(\dfrac{21}{8}\), \(\dfrac{5}{16}\)
(iii). \(-2\sqrt{3}\), \(-9\)
(iv). \(-\dfrac{3}{2\sqrt{5}}\), \(-\dfrac{1}{2}\)
(i) Polynomial: \(3x^2 + 8x + 4\); Zeroes: \(x=-2\), \(x=-\dfrac{2}{3}\).
(ii) Polynomial: \(16x^2 - 42x + 5\); Zeroes: \(x=\dfrac{5}{2}\), \(x=\dfrac{1}{8}\).
(iii) Polynomial: \(x^2 + 2\sqrt{3}x - 9\); Zeroes: \(x=\sqrt{3}\), \(x=-3\sqrt{3}\).
(iv) Polynomial: \(2\sqrt{5}x^2 + 3x - \sqrt{5}\); Zeroes: \(x=\dfrac{1}{\sqrt{5}}\), \(x=-\dfrac{\sqrt{5}}{2}\).
Key idea: If the sum of zeroes is \(S\) and the product is \(P\), then the monic quadratic is
\(x^2 - Sx + P = 0\).
To avoid fractions or surds in coefficients, we may multiply the whole equation by a suitable non-zero constant.
(i) \(S=-\dfrac{8}{3}\), \(P=\dfrac{4}{3}\).
Monic form: \(x^2 + \dfrac{8}{3}x + \dfrac{4}{3} = 0\).
Multiply by 3: \(3x^2 + 8x + 4 = 0\).
Factorise: \((3x+2)(x+2)=0\Rightarrow x=-\dfrac{2}{3},\,-2\).
(ii) \(S=\dfrac{21}{8}\), \(P=\dfrac{5}{16}\).
Monic form: \(x^2 - \dfrac{21}{8}x + \dfrac{5}{16} = 0\).
Multiply by 16: \(16x^2 - 42x + 5 = 0\).
Discriminant: \(\,42^2 - 4\cdot16\cdot5 = 1764 - 320 = 1444 = 38^2\).
Roots: \(x=\dfrac{42 \pm 38}{32} = \dfrac{80}{32},\,\dfrac{4}{32} = \dfrac{5}{2},\,\dfrac{1}{8}\).
(iii) \(S=-2\sqrt{3}\), \(P=-9\).
Monic form: \(x^2 + 2\sqrt{3}x - 9 = 0\).
\(x = \dfrac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4\cdot1\cdot(-9)}}{2}\)
\(= \dfrac{-2\sqrt{3} \pm \sqrt{12 + 36}}{2} = \dfrac{-2\sqrt{3} \pm \sqrt{48}}{2}\)
\(= \dfrac{-2\sqrt{3} \pm 4\sqrt{3}}{2} = \sqrt{3},\,-3\sqrt{3}\).
(iv) \(S=-\dfrac{3}{2\sqrt{5}}\), \(P=-\dfrac{1}{2}\).
Monic form: \(x^2 + \dfrac{3}{2\sqrt{5}}x - \dfrac{1}{2} = 0\).
Multiply by \(2\sqrt{5}\): \(2\sqrt{5}x^2 + 3x - \sqrt{5} = 0\).
Check zeroes by sum/product:
\(\dfrac{1}{\sqrt{5}} + \left(-\dfrac{\sqrt{5}}{2}\right) = -\dfrac{3}{2\sqrt{5}}\)
and \(\dfrac{1}{\sqrt{5}}\cdot\left(-\dfrac{\sqrt{5}}{2}\right) = -\dfrac{1}{2}\).