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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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