Question

find the zeros of the quadratic polynomial x square minus bracket root 3 + 1 bracket X + root 3 and verify the relationship between the zeros and the coefficient​

Answer 1

Let the polynomial be p(x).

⇒ p(x) = x² + (√3 + 1)x + √3

⇒ p(x) = x² + √3x + x + √3

⇒ p(x) = x(x + √3) + 1(x + √3)

⇒ p(x) = (x + 1)(x + √3)

⇒ 0 = x + 1 and 0 = x + √3

After each getting zero,

⇒ x = - 1 and x = - √3

According to relationship,

Sum of zeros = - Coefficient of x/Coefficient of x²

⇒ - 1 + (- √3)  = - (√3 + 1)

⇒ - (√3 + 1) = - (√3 + 1)

Product of zeros = Constant term/Coefficient of x²

⇒ - 1 × (- √3) = √3/1

⇒ √3 = √3

Hence relationship is verified.

Answer 2

Given Quadratic polynomial :-

x^2 - (√3 + 1) x + √3 = 0

X^2 - √3x - x + √3 = 0

Let's take " x " common from 1st two terms and " - 1 " common from the rest two terms

x (x - √3 ) - 1( x - √3 ) = 0

» Remember this - you need to get same terms in bracket in whatever the problem you do using factorization.

→ (x -1 ) (x - √3) = 0

Case 1 :-

x - 1 = 0

x = 1

Case 2 :-

x - √3 =0

x = √3

•°• we got the two zeroes of quatratic polynomial i.e 1 , √3

Let's verify the relationship between zeroes and coefficients if polynomial

The polynomial is the form of

ax^2 + bx + c

Here a = 1 , b = - (1+√3) ; c = √3

» product of the zeroes :-

1 × √3 = √3 -------(1)

We know that product of roots is

C/a = √3 /1 = √3---------(2)

Equation (1) = (2)

» Sum of the zeroes :-

1+ √3 = ( 1 +√3) ---------(3)

We know that Sum of roots is - b /a

= -[ -( 1 +√3)] = 1 + √3 ------(4)

Equation (3) = (4)

•°• The relationship between zeroes and coefficients is verified .

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