  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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