Question

find the find the zero of polynomial x square - 15 and verify the relation between zero and c o e f f i c i e n t s
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Answer 1

$\large\bf\underline{Given:-}$

• p(x) = x² - 15

$\large\bf\underline {To \: find:-}$

• zeroes of the given polynomial
• relationship between the zeroes and coefficients.

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

• »»p(x) = x² - 15

Finding zeroes of the given polynomial:-

➝ x² - 15 = 0

➝ x² = 15

➝ x = ±√15

• ≫ x = √15 or x = -√15

• » p(x) = x² - 15
• a = 1
• b = 0
• c = -15

$\large \star\underline{ \bf \dag \: Verification : - }$

Let α and β are the zeroes of the given polynomial.

• α = √15
• β = -√15

≫ Sum of zeroes = -b/a

➝ √15 +(-√15) = 0/1

➝ √15 - √15 = 0

➝ 0 = 0

≫ product of zeroes = c/a

➝ √15 × (-√15) = -15/1

➝ -15 = -15

So,

LHS = RHS

hence relationship is verified

Answer 2

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

$\sf{The \ zeroes \ of \ polynomial \ are \ -\sqrt15 \ and \sqrt15.}$

$\sf\orange{Given:}$

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

$\sf{\implies{x^{2}-15}}$

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

$\sf{The \ zeroes \ of \ the \ polynomial.}$

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

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

$\sf{\implies{x^{2}-15}}$

$\sf{\implies{x^{2}-\sqrt15^{2}}}$

$\sf{According \ to \ the \ identity}$

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

$\sf{\implies{(x+\sqrt15)(x-\sqrt15)}}$

$\sf{\implies{x=-\sqrt15 \ or \ \sqrt15}}$

$\sf\purple{\tt{\therefore{The \ zeroes \ of \ polynomial \ are \ -\sqrt15 \ and \sqrt15.}}}$

__________________________________

$\sf\blue{\underline{\underline{Verification:}}}$

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

$\sf{\implies{x^{2}-15}}$

$\sf{Here \ a=1, \ b=0 \ and \ c=-15}$

$\sf{Let \ the \ zeroes \ be \ \alpha \ and \ \beta}$

_______________________________

$\sf{\alpha+\beta=-\sqrt15+\sqrt15}$

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

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

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

$\sf{...from \ (1) \ and \ (2)}$

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

$\sf{\alpha\beta=-\sqrt15\times\sqrt15}$

$\sf{\therefore{\alpha\beta=-15...(3)}}$

$\sf{\frac{c}{a}=\frac{-15}{1}}$

$\sf{\therefore{\frac{c}{a}=-15...(4)}}$

$\sf{...from \ (3) \ and \ (4)}$

$\boxed{\sf{Product \ of \ zeroes=\frac{c}{a}}}$

$\sf{Hence, \ verified.}$