Question

alpha, beeta are the zereos of y2-12y+20, find qp whose zereos are 2alpha, 2beeta​

Answer 1

$\bold\blue{Question}$

$\bold{\alpha \ and \ \beta \ are \ the \ zeroes \ of}$

$\bold{y^{2}-12y+20, \ find \ quadratic \ polynomial}$

$\bold{whose \ zeroes \ are \ 2\alpha, \ 2\beta.}$

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{The \ required \ quadratic \ polynomial}$

$\bold{is \ y^{2}-24y+80.}$

$\bold\orange{Given:}$

$\bold{The \ given \ polynomial \ is}$

$\bold{=>y^{2}-12y+20}$

$\bold{=>Zeroes \ of \ polynomial \ are \ \alpha \ and \ \beta}$

$\bold\pink{To \ find:}$

$\bold{=>Quadratic \ polynomial \ with \ zeroes}$

$\bold{2\alpha \ and \ 2\beta.}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{The \ given \ quadratic \ polynomial \ is}$

$\bold{=>y^{2}-12y+20}$

$\bold{Here, \ a=1, \ b=-12, \ c=20.}$

$\bold{Sum \ of \ zeroes=\frac{-b}{a}}$

$\bold{\therefore{\alpha+\beta=\frac{-(-12)}{2}}}$

$\bold{\therefore{\alpha+\beta=12}}$

$\bold{Product \ of \ zeroes=\frac{c}{a}}$

$\bold{\therefore{\alpha×\beta=\frac{20}{1}}}$

$\bold{\therefore{\alpha×\beta=20}}$

________________________________

$\bold{Zeroes \ of \ new \ quadratic \ polynomial}$

$\bold{are \ 2\alpha \ and 2\beta.}$

$\bold{Sum \ of \ roots=2\alpha+2\beta}$

$\bold{\therefore{Sum \ of \ zeroes=2(\alpha+\beta)}}$

$\bold{\therefore{Sum \ of \ zeroes=2(12)}}$

$\bold{\therefore{Sum \ of \ zeroes=24}}$

$\bold{Product \ of \ zeroes=2\alpha×2\beta}$

$\bold{\therefore{Product \ of \ zeroes=2(\alpha×beta)}}$

$\bold{\therefore{Product \ of \ zeroes=4(\alpha×\beta)}}$

$\bold{\therefore{Product \ of \ zeroes=4(20)}}$

$\bold{\therefore{Product \ of \ zeroes=80}}$

$\bold{Quadratic \ polynomial \ is}$

$\bold{=>y^{2}-(Sum \ of \ zeroes)y+(Product \ of \ zeroes)}$

$\bold{=>y^{2}-24y+80}$

$\bold\purple{\tt{\therefore{The \ required \ quadratic \ polynomial}}}$

$\bold\purple{\tt{is \ y^{2}-24y+80.}}$