Question

find the zeros of each of the following quadratic polynomial and verify the relationship between the zero and their coefficient p(x) =x²-(√3+1)x+√3


pls helpthis​

Answer 1

★ Given :-

• p(x) =x²-(√3+1)x+√3

★ To Find :-

• The zeroes of the polynomial.
• The relationship between the zero and their coefficient.

★ Solution :-

$\sf \: x^2-(\sqrt{3}+1)x+\sqrt{3} = 0$

$\implies \sf \: \: x^2-\sqrt{3x} - x+\sqrt{3} = 0$

$\implies \sf \: \: x(x-\sqrt{3})- 1(x - \sqrt{3}) = 0$

$\implies \sf \: \: (x - \sqrt{3})(x - 1) = 0$

$\implies \sf \: \:( x - \sqrt{3} ) = 0$

$\implies \sf \: x = \sqrt{3}$

Or,

$\sf(x - 1) = 0$

$\implies \sf \: x = 1$

Therefore, zeroes of the polynomial are √3 and 1.

Relation between zeroes and coefficients:-

Sum of two zeroes :-

$\sf \sqrt{3} + 1 = - \dfrac{ ( - \sqrt{3} + 1 )}{1} = - \dfrac{b}{a}$

Product of two zeroes :-

$\sf \: \sqrt{3} \times 1 = \dfrac{ \sqrt{3} }{1} = \dfrac{c}{a}$

Answer 2

Answer

Given,

• p(x)= x^2 -(√3+1)x+√3

To find ,

• the zeros of the polynomial and the relationship between the zeros and their coefficient .

So,

$= > {x}^{2} - ( \sqrt{3} + 1)x + \sqrt{3} = 0$

$= > {x}^{2} - \sqrt{3} x - x + \sqrt{3} = 0$

$= > x(x - \sqrt{3} ) - 1(x - \sqrt{3} ) = 0$

$= > (x - \sqrt{3} )(x - 1) = 0$

$= > x - \sqrt{3} = 0$

$= > x = \sqrt{3}$

Or

$(x - 1) = 0 \\ = > x = 1$

so the zero of the polynomial is root 3 and 1 .

Now,

the relationship between the zero and their coefficient ;

sum of two zeroes

$\sqrt{3} + 1 = \frac{ - \sqrt{3} + 1 }{1} = \frac{ - b}{a}$

product of two zeros

$\sqrt{3} \times 1 = \frac{ \sqrt{3} }{1} = \frac{c}{a}$