Question

find the zeroes of the following x^2(√3+1)x+√3

pls give answer i will mark as brainliest ​

Answer 1

Answer:

Underroot 3 and 1

Step-by-step explanation:

Hi...☺

Here is your answer...✌

Given quadratic equation :

x² - (√3 + 1)x + √3 = 0

⇒ a = 1 , b = -(√3 + 1) , c = √3

Discriminant , D = b² - 4ac

D = (√3 + 1)² - 4×1×√3

D = (√3)² + 1² + 2√3 - 4√3

D = (√3)² + 1² - 2√3

D = (√3 - 1)²

Now using quadratic formula,

we get,

x = ( -b ± √D ) / 2a

x = [ √3 + 1 ± ( √3 - 1 ) ] / 2×1

x = (√3+1+√3-1)/2 or x = (√3+1-√3+1)/2

⇒ x = 2√3 / 2 or x = 2/2

⇒ x = √3 or x = 1

Answer 2

Answer:

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

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

=x(x+$\sqrt{3}$)+ 1(x+$\sqrt{3}$)=0                                  $\sqrt{3}$,1

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

ie, x+1=0 , x+$\sqrt{3}$=0

therefore, x= -1  ,     x= - $\sqrt{3}$

therefore the zeroes of the polynomial are -1 and -$\sqrt{3}$                                                      

Step-by-step explanation:

hope it helps................