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)     (ii)    (iii)        (iv)  Question

Solution

## We know that a quadratic polynomial whose sum and product of zeroes are given is (i) We have, sum of zeroes = $\frac{-8}{3}$ and product of zeroes = $\frac{4}{3}$ So, the required quadratic polynomial will be $f\left(x\right)=k\left({x}^{2}+\frac{8}{3}x+\frac{4}{3}\right)$ $f\left(x\right)=k\left({x}^{2}+\frac{8}{3}x+\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(3{x}^{2}+8x+4\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(3{x}^{2}+6x+2x+4\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(3x\left(x+2\right)+2\left(x+2\right)\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(3x+2\right)\left(x+2\right)$ Now, the zeroes are given by f(x) = 0. Thus,  (ii) We have, sum of zeroes = $\frac{21}{8}$ and product of zeroes = $\frac{5}{16}$ So, the required quadratic polynomial will be $f\left(x\right)=k\left({x}^{2}-\frac{21}{8}x+\frac{5}{16}\right)$ $f\left(x\right)=k\left({x}^{2}-\frac{21}{8}x+\frac{5}{16}\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{16}\left(16{x}^{2}-42x+5\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{16}\left(16{x}^{2}-40x-2x+5\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{16}\left(16{x}^{2}-2x-40x+5\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(2x\left(8x-1\right)-5\left(8x-1\right)\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{3}\left(8x-1\right)\left(2x-5\right)$ Now, the zeroes are given by f(x) = 0. Thus,  (iii) We have, sum of zeroes = $-2\sqrt{3}$ and product of zeroes = −9. So, the required quadratic polynomial will be $f\left(x\right)=k\left({x}^{2}+2\sqrt{3}x-9\right)$. $f\left(x\right)=k\left({x}^{2}+2\sqrt{3}x-9\right)\phantom{\rule{0ex}{0ex}}=k\left({x}^{2}+3\sqrt{3}x-\sqrt{3}x-9\right)\phantom{\rule{0ex}{0ex}}=k\left(x+3\sqrt{3}\right)\left(x-\sqrt{3}\right)$ Now, the zeroes are given by f(x) = 0. Thus, . (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 $f\left(x\right)=k\left({x}^{2}+\frac{3}{2\sqrt{5}}x-\frac{1}{2}\right)$. $f\left(x\right)=\frac{k}{2\sqrt{5}}\left(2\sqrt{5}{x}^{2}+3x-\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{2\sqrt{5}}\left(2\sqrt{5}{x}^{2}+5x-2x-\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{2\sqrt{5}}\left[\sqrt{5}x\left(2x+\sqrt{5}\right)-1\left(2x+\sqrt{5}x\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{k}{2\sqrt{5}}\left(2x+\sqrt{5}\right)\left(\sqrt{5}x-1\right)$ Now, the zeroes are given by f(x) = 0. Thus, .MathematicsRD Sharma (2016)Standard X

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