Question

Find the zeros of the quadratic polynomial x square minus 2 and verify the relationship between the zeros and the coefficients
plz tell

Answer 1

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

$\bold{Zeroes \ are \ -\sqrt2 \ and \ \sqrt2}$

$\bold\orange{Given:}$

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

$\bold{=>x^{2}-2}$

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

$\bold{Zeroes \ of \ polynomial.}$

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

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

$\bold{=>x^{2}-2}$

$\bold{Here, \ a=1, \ b=0, \ and \ c=-2}$

$\bold{By \ identity}$

$\bold{a^{2}-b^{2}=(a+b)(a-b)}$

$\bold{=>x^{2}-(\sqrt2)^{2}}$

$\bold{=>(x+\sqrt2)(x-\sqrt2)}$

$\bold{\therefore{x=-\sqrt2 \ or \ \sqrt2}}$

$\bold\purple{\tt{\therefore{Zeroes \ are \ -\sqrt2 \ and \ \sqrt2}}}$

_________________________________

$\bold\blue{Verification}$

$\bold{Let \ \alpha \ be -\sqrt2 \ and \ \beta \ be \ \sqrt2}$

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

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

_______________________________

$\bold{\alpha+\beta=-\sqrt2+\sqrt2}$

$\bold{\alpha+\beta=0...(1)}$

$\bold{\frac{-b}{a}=\frac{0}{1}}$

$\bold{\frac{-b}{a}=0...(2)}$

$\bold{from \ (1) \ and \ (2)}$

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

________________________________

$\bold{\alpha×\beta=-\sqrt2×\sqrt2}$

$\bold{\alpha×\beta=-2...(3)}$

$\bold{\frac{c}{a}=\frac{-2}{1}}$

$\bold{\frac{c}{a}=-2...(4)}$

$\bold{from \ (3) \ and (4)}$

$\bold{Product \ of \ zeroes=-2}$

$\bold{Hence, \ verified.}$