Question

Zeroes of the polynomial p(x) = x^2 - 9 are ​

Answer 1

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{Zeroes \ of \ the \ polynomial \ are}$

$\sf{-3 \ and \ 3.}$

$\sf\orange{Given:}$

$\sf{\implies{p(x)=x^{2}-9}}$

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

$\sf{Zeroes \ of \ the \ polynomial.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{\implies{x^{2}-9}}$

$\sf{\implies{x^{2}-3^{2}}}$

$\sf{According \ to \ identity}$

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

$\sf{\implies{(x+3)(x-3)}}$

$\sf{\implies{\therefore{x=-3 \ or \ 3}}}$

$\sf\purple{\tt{\therefore{Zeroes \ of \ the \ polynomial \ are}}}$

$\sf\purple{\tt{-3 \ and \ 3.}}$

Answer 2

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

$\green{\tt{\therefore{x=\pm 3}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {x}^{2} - 9 = 0 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies zeroes \: of \: polynomial =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {x}^{2} - 9 = 0 \\ \\ \tt: \implies {x}^{2} = 9 \\ \\ \tt: \implies x = \sqrt{9} \\ \\ \green{\tt: \implies x = \pm3} \\ \\ \bold{Alternate \: method : } \\ \tt: \implies {x}^{2} - 9 = 0 \\ \\ \tt \circ \: a = 1 \: \: \: \: \: b = 0 \: \: \: \: \: c = - 9\\ \\ \tt: \implies x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \tt: \implies x = \frac{ - 0 \pm \sqrt{ {0}^{2} - 4 \times 1 \times ( - 9) } }{2 \times 1} \\ \\ \tt: \implies x = \frac{ \pm \sqrt{36} }{2} \\ \\ \tt: \implies x = \frac{ \pm6}{2} \\ \\ \green{\tt: \implies x = \pm 3}$