Question

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 factorization.
(i) $-\frac{8}{3},\frac{4}{3}$ (ii) $\frac{21}{8},\frac{5}{16}$ (iii) $-2\sqrt{3},-9$ (iv) $\frac{-3}{2\sqrt{5}},-\frac{1}{2}$Question

Answer 1

We know that a quadratic polynomial whose sum and product of zeroes are given is
$f\left(x\right)=k\left\{x^2-\left(\mathrm{Sum}\mathrm{of}\mathrm{zeroes}\right)x+\mathrm{Product}\mathrm{of}\mathrm{zeroes}\right\}$

(i) We have, sum of zeroes = $\frac{-8}{3}$ and product of zeroes = $\frac{4}{3}$
So, the required quadratic polynomial will be $f\left(x\right)=k\left(x^2+\frac{8}{3}x+\frac{4}{3}\right)$
$$\begin{gathered} f\left(x\right)=k\left(x^2+\frac{8}{3}x+\frac{4}{3}\right) \\ =\frac{k}{3}\left(3x^2+8x+4\right) \\ =\frac{k}{3}\left(3x^2+6x+2x+4\right) \\ =\frac{k}{3}\left(3x\left(x+2\right)+2\left(x+2\right)\right) \\ =\frac{k}{3}\left(3x+2\right)\left(x+2\right) \end{gathered}$$
Now, the zeroes are given by f(x) = 0.
Thus, $x=-\frac{2}{3}\mathrm{and}x=-2$
(ii) We have, sum of zeroes = $\frac{21}{8}$ and product of zeroes = $\frac{5}{16}$
So, the required quadratic polynomial will be $f\left(x\right)=k\left(x^2-\frac{21}{8}x+\frac{5}{16}\right)$
$$\begin{gathered} f\left(x\right)=k\left(x^2-\frac{21}{8}x+\frac{5}{16}\right) \\ =\frac{k}{16}\left(16x^2-42x+5\right) \\ =\frac{k}{16}\left(16x^2-40x-2x+5\right) \\ =\frac{k}{16}\left(16x^2-2x-40x+5\right) \\ =\frac{k}{3}\left(2x\left(8x-1\right)-5\left(8x-1\right)\right) \\ =\frac{k}{3}\left(8x-1\right)\left(2x-5\right) \end{gathered}$$
Now, the zeroes are given by f(x) = 0.
Thus, $x=\frac{1}{8}\mathrm{and}x=\frac{5}{2}$
(iii) We have, sum of zeroes = $-2\sqrt{3}$ and product of zeroes = −9.
So, the required quadratic polynomial will be $f\left(x\right)=k\left(x^2+2\sqrt{3}x-9\right)$.
$$\begin{gathered} f\left(x\right)=k\left(x^2+2\sqrt{3}x-9\right) \\ =k\left(x^2+3\sqrt{3}x-\sqrt{3}x-9\right) \\ =k\left(x+3\sqrt{3}\right)\left(x-\sqrt{3}\right) \end{gathered}$$
Now, the zeroes are given by f(x) = 0.
Thus, $x=-3\sqrt{3}\mathrm{and}x=\sqrt{3}$.

(iv) We have, sum of zeroes = $\frac{-3}{2\sqrt{5}}$ and product of zeroes = $\frac{-1}{2}$
So, the required quadratic polynomial will be $f\left(x\right)=k\left(x^2+\frac{3}{2\sqrt{5}}x-\frac{1}{2}\right)$.
$$\begin{gathered} f\left(x\right)=\frac{k}{2\sqrt{5}}\left(2\sqrt{5}{x}^2+3x-\sqrt{5}\right) \\ =\frac{k}{2\sqrt{5}}\left(2\sqrt{5}{x}^2+5x-2x-\sqrt{5}\right) \\ =\frac{k}{2\sqrt{5}}\left[\sqrt{5}x\left(2x+\sqrt{5}\right)-1\left(2x+\sqrt{5}x\right)\right] \\ =\frac{k}{2\sqrt{5}}\left(2x+\sqrt{5}\right)\left(\sqrt{5}x-1\right) \end{gathered}$$

Now, the zeroes are given by f(x) = 0.
Thus, $x=\frac{-\sqrt{5}}{2}\mathrm{and}x=\frac{1}{\sqrt{5}}$.