Question

Find the zeros of x² square minus 7 and verify the relationship between the zeros and its coefficient

Answer 1

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

• ✦ p(x) = x² - 7

$\\\large{\underline{\bf{\blue{To\:Find:-}}}}$

• ✦ we need to find the zeroes of p(x) and also find the relationship between the zeroes and coefficients.

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

$\\\\ \longmapsto \rm\: {x}^{2} - 7 \\ \\\longmapsto \bf \green{\: {x }^{2} - {y}^{2} = (x + y)(x - y)} \\ \\ \longmapsto \rm\: {x}^{2} - (\sqrt{7}) {}^{2} \: \\ \\ \longmapsto \rm\:(x + \sqrt{7})(x - \sqrt{7}) \\ \\\longmapsto \bf\: x = - \sqrt{7} \: \: or \: \: x = \sqrt{7}$

Now ,

Relationship between the zeroes and coefficients:-

• Let α = -√7
• β = √7

sum of zeroes = - b/a

α + β = - b/a

$\longmapsto \rm\: - \sqrt{7} + \sqrt{7} = \frac{0}{1} \: \\ \\ \longmapsto \rm0 = 0$

product of zeroes = c/a

αβ = c/a

$\longmapsto \rm\: - \sqrt{7} \times \sqrt{7} = \frac{ - 7}{1} \: \\ \\\longmapsto \rm\: - 7 = - 7$

LHS = RHS

hence Relationship is verified

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Answer 2

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

$\sf{Zeroes \ are \ -\sqrt7 \ and \ \sqrt7}$

$\huge\purple{Given:}$

$\sf{p(x)=x^{2}-7}$

$\huge\blue{To \ find:}$

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

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

$\sf{\implies{x^{2}-7}}$

$\sf{\implies{x^{2}-\sqrt7^{2}}}$

$\sf{By \ identity}$

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

$\sf{\implies{(x+\sqrt7)(x-\sqrt7)}}$

$\sf{\therefore{x=-\sqrt7 \ or \ \sqrt7}}$

$\sf\purple{\tt{\therefore{Zeroes \ are \ -\sqrt7 \ and \ \sqrt7}}}$

$\sf\blue{Verification:}$

$\sf{\implies{p(x)=x^{2}-7}}$

$\sf{Here, \ a=1, \ b=0 \ and \ c=-7.}$

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

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

_______________________________

$\sf{Let \ \alpha \ be \ -\sqrt7 \ and \ \beta \ be \ \sqrt7}$

$\sf{\alpha+\beta=-\sqrt7+\sqrt7}$

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

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

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

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

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

_______________________________

$\sf{\alpha×\beta=-\sqrt7×\sqrt7}$

$\sf{\therefore{\alpha×\beta=-7...(3)}}$

$\sf{\frac{c}{a}=\frac{-7}{1}}$

$\sf{\therefore{\frac{c}{a}=-7...(4)}}$

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

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

$\sf{Hence, \ verified.}$