Question

find the zeroes of the polynomial 3x^2-2 and verify the relationship between the zeroes and the coefficients.

Answer 1

Answer:

x = 4 or -4

Step-by-step explanation:

Given polynomial, f(x) = 3x² - 2

Let f(x) = 0 (to find the zeroes)

Here, comparing it with a standard quadratic equation, ax² + bx + c = 0,

we get a = 3; b = 0; c = -2.

Using quadratic formula (derivation in link in comments), we get

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\\\= \dfrac{-(0) \pm \sqrt{(0)^2 - 4(3)(-2)}}{2(3)}\\\\\\= \dfrac{ \pm \sqrt{24}}{6}\\\\\\= \pm 2 \sqrt{6} \div 6 = \pm \dfrac{\sqrt{6}}{3}$

∴ x = √6/3 or -√6/3

Verification of relationship between zeroes and the coefficients:-

(1) Sum of zeroes = -√6/3 + √6/3 = 0

and,

-b/a = -(0)/3 = 0/(-3) = 0

∴ Sum of zeroes = -b/a = 0

(2) Product of zeroes = -√6/3 * √6/3 = -6/9 = -2/3

and,

c/a = -2/3

∴ Product of zeroes = c/a

Hence proved!

Answer 2

GIVEN :

Polynomial : 3x² - 2 = 0

SOLUTION :

3x² - 2 = 0

=> 3x² = 2

=> x² = 2/3

=> x = ± √( 2/3 )

Hence,

Hence,Roots of the Polynomial are -√(2/3) and √(2/3)

VERIFICATION:-

Sum of Zeroes :

= α + β

= √2/3 + (- √2/3)

= √2/3 - √2/3

= 0

Using Coefficients :

= α + β

= -b/a

= -0/3

= 0

Product of Zeroes :

= αβ

= √2/3 × [ - √(2/3)

= -2/3

Using Coefficients :

= αβ

= c/a

= -2/3

Hence Verified!