Question

find zeroes of polynomial x²-5 and verify relationship between zeroes and its coefficients​

Answer 1

Step-by-step explanation:

x^2-√5^2

=(x-√5)(x+√5) a^2-b^2=(a-b)(a+b)

Roots:

x-√5=0

x=√5

x+√5=0

x=-√5

Verification:

sum of zeros=-b/a

=0

-0/1=0

product of zeros=c/a

5/1=-5

√5(-√5) =-5

Hence verified

Answer 2

$\bold{\underline{\sf{\red{Given\: Polynomial}}}}$

$:\implies\sf\:x^2 - 5$

$:\implies\sf\:x = \sqrt{5} \: and \: - \sqrt{5}$

Zeroes of the given polynomial are $\sf\: \sqrt{5} \: and \: - \sqrt{5}$

$\bold{\underline{\sf{\red{Now,\:Verification}}}}$

$:\implies\sf\: (\alpha \:+\: \beta) = \dfrac{-b}{a}$

$:\implies\sf\: (\alpha \:+\:\beta) = \sqrt{5} \:+\: - \sqrt{5}$

$:\implies\sf\: 0$

Now, product of Zeroes:

$:\implies\sf\: (\alpha \:\beta) = \dfrac{c}{a}$

$:\implies\sf\: \dfrac{-5}{1} = - 5$

$:\implies\sf\: \sqrt{5} \: \times ( - \sqrt{5} )= - 5$

$\bold{\underline{\sf{\blue{Hence,\:Verified!}}}}$