Question

find the zeros of the quadratic polynomial x^2-2
and verify the relationship between the zeroes and the co efficient ​

Answer 1

AnswEr:-

❀ The roots of the given equation are $\tt \sqrt{2}\:\:And\:\: -\sqrt{2}.$

ExplanaTion:-

Given quadratic polynomial:-

• $\tt\longrightarrow x^2-2$.

To Find:-

• The roots of the given polynomial and also we have to verify the relationship between zeroes and the Coefficient.

Now,

To Find out the roots the given polynomial should be equal to zeroe.

$\\ \tt\mapsto {x}^{2} - 2 = 0. \\ \\ \tt\mapsto {x}^{2} = 2. \\ \\ \tt\mapsto \fbox{\underline{\underline{x = \pm \: \sqrt{2} .}}}$

$\therefore$ The required values of x are $\tt\sqrt{2}\:\:And\:\: -\sqrt{2}.$

$\rule{200}2$

VerificaTion:-

Here,

$\\ \tt\longrightarrow {x}^{2} - 2 \: \: \rm can \: \: be \: \: written \: \: as \\ \\ \tt\longrightarrow {x}^{2} + 0x - 2.$

Now we know that:-

$\\ \\\rm{\underline{\underline{Sum\:\:Of\:\:Zeroes:-}}}\\ \\ \tt =\dfrac{-Coefficient\:\:Of\:\:x}{Coefficient\:\:Of\:\:x^2} [\dfrac{-b}{a}].\\ \\ \tt= \dfrac{0}{1}.\\ \\ \tt = 0.$

Also,

$\\ \\ \tt\leadsto \sqrt{2} + ( - \sqrt{2}) .\\ \\ \tt = \sqrt{2} - \sqrt{2} . \\ \\ \tt = 0.$

$\tt\therefore$ Sum of zeroes = $\tt\dfrac{-b}{a}.$

Similarly,

$\\ \\ \rm{\underline{\underline{Product\:\:Of\:\:Zeroes:-}}}\\ \\ \tt =\dfrac{Constant\:\:Term}{Coefficient\:\:Of\:\:x^2} [\dfrac{c}{a}].\\ \\ \tt= \dfrac{-2}{1}.\\ \\ \tt = -2.$

Also,

$\\ \\ \tt\leadsto \sqrt{2} \times ( - \sqrt{2} ) \\ \\ \tt = \sqrt{2} \times - \sqrt{2} . \\ \\ \tt = - 2.$

$\tt\therefore$ Product of zeroes = $\tt\dfrac{c}{a}.$

Hence Verified.