Question

find the zeros of the quadratic polynomial 3x2-2 verify the relationship between the zeros and the cofficients
​

Answer 1

Step-by-step explanation:

Given -

• p(x) = 3x² - 2

To Find -

• Zeroes of the polynomial and verify the relationship between the zeroes and the coefficient.

Now,

→ 3x² - 2

→ (√3x)² - (√2)²

Now, It is in the form of

• a² - b² = (a + b)(a - b)

→ (√3x - √2)(√3x + √2)

Zeroes are -

→ √3x - √2 = 0 and √3x + √2 = 0

→ x = √2/√3 and x = -√2/√3

Verification :-

• α + β = -b/a

→ -√2/√3 + √2/√3 = -(0)/3

→ 0 = 0

LHS = RHS

And

• αβ = c/a

→ -√2/√3 × √2/√3 = -2/3

→ -2/3 = -2/3

LHS = RHS

Hence,

Verified...

Answer 2

Hyy Dude ♡

Answer:

-√(2/3),√(2/3) are zeroes of the given polynomial

Explanation:

Given Quadratic polynomial p(x) = 3x²-2

i) To find the zeroes of p(x),

we must take p(x)=0

• 3x²-2 = 0

• => 3x² = 2

• => x² = 2/3

• => x = ±√(2/3)

Verification:

Let m,n are two zeroes of the polyomial,

Let m,n are two zeroes of the polyomial,m = -√(2/√3)and n = √(2/3)

Compare p(x) with ax²+bx+c , we get

a = 3 , b = 0 , c=-2

i) sum of the zeroes =

• = -√(2/3)+√(2/3)

• = 0

= - x - coefficient / x² coefficient

ii) Product of the zeros

• = [-√(2/3)][√(2/3)]

• = -2/3

Constant / x² - coefficient

Hope it's helps you ☺️