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# 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.

$ \frac{-8}{3}, \frac{4}{3} $

**To do:**

Here, we have to find the quadratic polynomials whose sum and product of the zeros are as given.

**Solution:**

(i) Sum of the zeros of a polynomial$=-\frac{8}{3}$.

Product of the zeros of the polynomial$=\frac{4}{3}$.

A quadratic polynomial formed for the given sum and product of zeros is given by:

$f(x) = x^2 -( \text { sum of zeros }) x + ( \text { product of zeros })$

Therefore,

The required polynomial f(x) is,

$x^2- (-\frac{8}{3})x + (\frac{4}{3})$

$=x^2 + \frac{8}{3}x + \frac{4}{3}$

To find the zeros of f(x), we put $f(x) = 0$.

This implies,

$x^2 + \frac{8}{3}x + \frac{4}{3} = 0$

Multiplying by 3 on both sides, we get,

$3(x^2) + 3(\frac{8}{3})x + 3(\frac{4}{3})= 0$

$3x^2+8x+4=0$

$3x^2 + 6x + 2x + 4 = 0$

$3x(x + 2) + 2(x + 2) = 0$

$(x + 2) (3x + 2) = 0$

$(x + 2) = 0$ and $(3x + 2) = 0$

$x=-2$ and $x=-\frac{2}{3}$

Therefore, the two zeros of the quadratic polynomial are $-2$ and $-\frac{2}{3}$.

(ii) Sum of the zeros of a polynomial$=\frac{21}{8}$.

Product of the zeros of the polynomial$=\frac{5}{16}$.

A quadratic polynomial formed for the given sum and product of zeros is given by:

$f(x) = x^2 -( \text { sum of zeros }) x + ( \text { product of zeros })$

Therefore,

The required polynomial f(x) is,

$x^2- (\frac{21}{8})x + (\frac{5}{16})$

$=x^2 - \frac{21}{8}x + \frac{5}{16}$

To find the zeros of f(x), we put $f(x) = 0$.

This implies,

$x^2 -\frac{21}{8}x + \frac{5}{16} = 0$

Multiplying by 16 on both sides, we get,

$16(x^2) -16(\frac{21}{8})x + 16(\frac{5}{16})= 0$

$16x^2-42x+5=0$

$16x^2 -40x - 2x + 5 = 0$

$8x(2x - 5) - 1(2x -5) = 0$

$(2x - 5) (8x-1) = 0$

$(2x -5) = 0$ and $(8x -1) = 0$

$x=\frac{5}{2}$ and $x=\frac{1}{8}$

Therefore, the two zeros of the quadratic polynomial are $\frac{5}{2}$ and $\frac{1}{8}$.

(iii) Sum of the zeros of a polynomial$=-2\sqrt{3}$.

Product of the zeros of the polynomial$=-9$.

A quadratic polynomial formed for the given sum and product of zeros is given by:

$f(x) = x^2 -( \text { sum of zeros }) x + ( \text { product of zeros })$

Therefore,

The required polynomial f(x) is,

$x^2- (-2\sqrt{3})x + (-9)$

$=x^2 +2\sqrt{3}x -9$

To find the zeros of f(x), we put $f(x) = 0$.

This implies,

$x^2 +2 \sqrt{3}x -9 = 0$

$x^2 +3\sqrt{3} x -\sqrt{3}x -9 = 0$

$x(x + 3\sqrt{3}) -\sqrt{3}(x +3\sqrt{3}) = 0$

$(x + 3\sqrt{3}) (x -\sqrt{3}) = 0$

$(x + 3\sqrt{3}) = 0$ and $(x -\sqrt{3}) = 0$

$x=-3\sqrt{3}$ and $x=\sqrt{3}$

Therefore, the two zeros of the quadratic polynomial are $-3\sqrt{3}$ and $\sqrt{3}$.

(iv) Sum of the zeros of a polynomial$=-\frac{3}{2\sqrt5}$.

Product of the zeros of the polynomial$=-\frac{1}{2}$.

A quadratic polynomial formed for the given sum and product of zeros is given by:

$f(x) = x^2 -( \text { sum of zeros }) x + ( \text { product of zeros })$

Therefore,

The required polynomial f(x) is,

$x^2- (-\frac{3}{2\sqrt5})x + (-\frac{1}{2})$

$=x^2 + \frac{3}{2\sqrt5}x - \frac{1}{2}$

To find the zeros of f(x), we put $f(x) = 0$.

This implies,

$x^2 + \frac{3}{2\sqrt5}x - \frac{1}{2} = 0$

Multiplying by $2\sqrt5$ on both sides, we get,

$2\sqrt5(x^2) +2\sqrt5(\frac{3}{2\sqrt5})x - 2\sqrt5(\frac{1}{2})= 0$

$2\sqrt5x^2+3x-\sqrt5=0$

$2\sqrt5x^2 + 5x - 2x -\sqrt5= 0$

$\sqrt5x(2x + \sqrt5) -1(2x +\sqrt5) = 0$

$(2x +\sqrt5) (\sqrt5x - 1) = 0$

$(2x +\sqrt5) = 0$ and $(\sqrt5x - 1) = 0$

$2x=-\sqrt5$ and $\sqrt5x=1$

$x=\frac{-\sqrt5}{2}$ and $x=\frac{1}{\sqrt5}$

Therefore, the two zeros of the quadratic polynomial are $\frac{-\sqrt5}{2}$ and $\frac{1}{\sqrt5}$.

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