Question

If alpha and beta are the zeroes of the polynomial 6y(squared)+7y+2 , find quadratic polynomial , whose zeroes are 1/ alpha and 1/ beta ?

Answer 1

hope this helps you

Answer 2

$\underline{\bf \: Solution:-}}}$

$\rm \: The \: given \: polynomial \: is \: \\\rm \: \: \: \: 6y^{2} -7y+2$

$\rm \: Comparing \: it \: with \: ay^{2} +by+c, \: we \: get$

$\rm \: a=6, \: b=-7, \: c=2$

$\therefore \: \rm \: Sum \: of \: zeroes=\alpha +\beta = - \frac{b}{a}=\frac{7}{6}$

$\rm \: \implies \: \alpha +\beta =\frac{7}{6} \: \: \: \: \: \: \: \: \: \: ...(1)$

$\rm \: and, \: product \: of \: zeroes = \alpha \beta =\frac{c}{a} =\frac{2}{6}=\frac{1}{3}$

$\implies \: \rm \: \alpha \beta =\frac{1}{3} \: \: \: \: \: \: \: \: \: \: ...(2)$

$\bf \: For \: a \: quadratic \: polynomial \: whose\\\bf \: zeroes \: are \: \frac{1}{\alpha} \: and \frac{1}{\beta}$

$\rm \: S=Sum \: of \: zeroes$

$\rm \: = \frac{1}{\alpha}+ \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta }= \frac{7/6}{1/3} \: \: \: \: \: \: \: \: \: \:[ From \: (1) \: and \: (2)]$

$\rm = \frac{7}{6} \times \: \frac{3}{1}=\frac{7}{2}$

$\rm \: P=Product \: of \: zeroes$

$\rm \: =\frac{1}{\alpha} \frac{1}{\beta}=\frac{1}{\alpha \beta}=\frac{1}{\bigg( \frac{1}{3} \bigg)} \: \: \: \: \: \: \: \: \: \: [From \: (2)]$

$\rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: 3$

$\therefore \: \rm \: The \: required \: quadratic \: polynomial \: is$

$\rm \: k (x^{2} -Sx+P) \: where \: k\neq \: 0 \: is \: real$

$\rm \: or, \: k \: \bigg(x^{2} - \frac{7}{2}x+3 \bigg)=\frac{k}{2}(2x^{2} -7x+6)$

$\rm \: Taking \: k=2, one \: such \: quadratic \: polynomial\\\rm \: is \: \frac{2}{2}(2x^{2} -7x+6)=2x^{2} -7x+6.$