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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.

(i) $\frac{-8}{3},\frac{4}{3}$

(ii) $\frac{21}{8},\frac{5}{16}$

(iii) $-2\sqrt{3}, -9$

(iv) $\frac{-3}{2\sqrt{5}},-\frac{1}{2}$

Answers (1)

(i) Answer. $\left [ -2 , -\frac{2}{3}\right ]$

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

Here, sum of zeroes $=\frac{-8}{3}$

Product of zeroes $=\frac{4}{3}$

Let p(x) is the required polynomial

p(x) = x² – (sum of zeroes)x + (product of zeroes)

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

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

Multiply by 3

$p(x) = 3x^2 + 8x + 4$

Hence, $3x^2 + 8x + 4$ is the required polynomial

$p(x) = 3x^2 + 8x + 4$

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

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

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

$x = -2,\frac{2}{3}$ are the zeroes of p(x).

(ii) Answer. $\left [\frac{5}{2}, \frac{1}{8} \right ]$

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

Here, sum of zeroes $=\frac{21}{8}$

Product of zeroes $=\frac{5}{16}$

Let p(x) is the required polynomial

p(x) = x² – (sum of zeroes)x + (product of zeroes)

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

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

Multiply by 16 we get

$p(x) = 16x^2 - 42x + 5$

Hence, $16x^2 - 42x + 5$ is the required polynomial

$p(x) = 16x^2 - 42x + 5$

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

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

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

$x=\frac{5}{2}, \frac{1}{8}$ are the zeroes of p(x).

(iii) Answer. $\left [-3\sqrt{3}, \sqrt{3} \right ]$

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

Here, sum of zeroes $-2\sqrt{3}$

Product of zeroes = –9

Let p(x) is required polynomial

p(x) = x² – (sum of zeroes)x + (product of zeroes)

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

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

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

Hence, $x^{2 } +2\sqrt{3}x+(-9)$ is the required polynomial

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

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

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

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

$x =-3\sqrt{3}, \sqrt{3}$ are the zeroes of p(x)

(iv) Answer. $\left [\frac{\sqrt{5}}{2}, \frac{-1}{\sqrt{5}} \right ]$

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

Here sum of zeroes $\frac{-3}{2\sqrt{5}}$

Product of zeroes $=-\frac{1}{2}$

Let p(x) is the required polynomial

p(x) = x² – (sum of zeroes)x + (product of zeros)

$p(x)=x^{2}-\left (\frac{-3}{2\sqrt{5}} \right )x+ \frac{-1}{2}$

$p(x)=x^{2}+\frac{3}{2\sqrt{5}} x- \frac{1}{2}$

Multiplying by $2 \sqrt{5}$ we get

$p(x)=2 \sqrt{5}x^{2}+3x-\sqrt{5}$

Hence, $2 \sqrt{5}x^{2}+3x-\sqrt{5}$ is the required polynomial

$p(x)=2 \sqrt{5}x^{2}+3x-\sqrt{5}$

$=2 \sqrt{5}x^{2}+5x-2x -\sqrt{5}$

= $\sqrt{5}(2x+\sqrt{5})-1(2x+\sqrt{5})$

= $(\sqrt{5}x -1)(2x+\sqrt{5})$

$\left [\frac{-\sqrt{5}}{2}, \frac{1}{\sqrt{5}} \right ]$ are the zeroes of p(x).

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