Question

find the zeroes of the quadratic polynomial x^2-16 and verify the relationship between zeroes and coefficients

Answer 1

Explanation:

Zeroes of polynomial = 4 & 12

Given :- Polynomial:-

☛ x² - 16x + 48

Let the zeroes be α & β

⇒ x² - 16x + 48

⇒ x² - 4x - 12x + 48

⇒ x(x - 4) - 12(x - 4)

⇒ (x - 4)(x - 12)

↠ x = 4 |or| x = 12

∴ α = 4

∴ β = 12

Therefore,

a = 1

b = -16

c = 48

Verifying the relationship between zeroes & coefficients.

Relationship 1:-

☛ Sum of zeroes = -b/a

↠ (α + β) = -b/a

↠ 4 + 12 = -(-16)/1

↠ 16 = 16/1

↠ 16 = 16 [Verified!]

Relationship 2:-

☛ Product of zeroes = c/a

↠ αβ = 48/1

↠ 4 × 12 = 48

↠ 48 = 48 [Verified!]

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Answer 2

The sum of the zeros is 0 and the product of zeros -16

Given:

The quadratic polynomial  $x^{2} -16$ which is an equation

To Find:

The sum of zeros and product of zeros of the given polynomial

Solution:

The equation is $x^{2} -16$

a$x^{2}$ +bx + c = 0

a = 1

Solving the equation ,

We get  ,

x = 4, x = -4

Which gives α = 4,  β = -4

Sum of the roots is given by,

α +  β = $\frac{-b}{a}$

4-4  =  $\frac{(0)}{1}$

LHS = RHS

α +  β =  0

Product of the rots is given by,

(α)( β ) = $\frac{c}{a}$

-16 = $\frac{-16}{1}$

LHS = RHS

(α)( β ) = -16

Hence, sum of the roots is 0 and the product of roots is -16

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