Question

find the polynomilas of
$\sqrt{3} - \sqrt{3}$
​

Answer 1

Answer:

Step-by-step explanation:

3+1/root3+1/3+root3+1/root3-3

3 + 1 \div  \sqrt{3}  + 1 \div 3 +  \sqrt{3}  + 1 \div  \sqrt{3}  - 3 =  

=>3+1/3-3+1/root3+1/root3+root3

=>1/3+2/root3+root3

=>1/3+2+(root3)^2/root3

=>1/3+2+3/root3

=>1/3+5/root3

=>root3+15/3root3

=  >  \sqrt{3}  + 15 \div 3 \sqrt{3}  

Follow me....

Answer 2

$\sf\large\underline{Correct\: Question:}$

Find the quardric polynomial whose zeroes are √3 and -√3.

$\sf\large\underline{Given:}$

• $\rm{The\: zeroes\:are\:\sqrt{3}\:and\:-\sqrt{3}}$

$\sf\large\underline{To\: Find:}$

• $\rm{Quardric\: polynomial}$

$\sf\large\underline{Solution:}$

• At first calculate sum of 2 zeroes and product of to zeroes than calculate quardric polynomial by using formula]

$\sf\large\underline{Formula\: used:}$

$\rm{\implies Sum\:of\:2\: zeroes=\alpha+\beta}$

$\rm{\implies Sum\:of\:2\: zeroes=\sqrt{3}+(-\sqrt{3}}$

$\rm{\implies Sum\:of\:2\: zeroes=\sqrt{3}-\sqrt{3}}$

$\rm{\implies Sum\:of\:2\: zeroes=0}$

• Now, Find out product of 2 zeroes]

$\rm{\implies Product\:of\:2\: zeroes=\alpha\times\:\beta}$

$\rm{\implies Product\:of\:2\: zeroes=\sqrt{3}\times\:(-\sqrt{3}}$

$\rm{\implies Product\:of\:2\: zeroes=-\sqrt{9}}$

$\rm{\implies Product\:of\:2\: zeroes=-3}$

• Now, calculate quardric polynomial here]

$\rm{\implies x^2-(sum\:of\:2\: zeroes)x+(product\:of\:2\: zeroes)}$

$\rm{\implies x^2-0\times\:x+(-3)}$

$\rm{\implies x^2-0-3}$

$\rm{\implies x^2-3}$

$\sf\large{Hence,}$

$\rm{\implies Quardric\: polynomial=x^2-3}$