Question

by method of completing square x2-(√3+1)x+√3=0​

Answer 1

Step-by-step explanation:

block of the periodic table is a set of elements unified by the orbitals their valence electrons or vacancies lie in.[1] The term appears to have been first used by Charles Janet.[2] Each block is named after its characteristic orbital: s-block, p-block, d-block, and f-block.

A long periodic table showing, from left to right: the s-, d-, f-, and p-blocks. The f-block, normally shown as a footnote, here splits the d-block into two. While this splitting is the more common form in the literature, a minority advocates placing the f-block between the s- and d-blocks.[1]

The block names (s, p, d, and f) are derived from the spectroscopic notation for the value of an electron's azimuthal quantum number: shape (0), principal (1), diffuse (2), or fundamental (3). Succeeding notations proceed in alphabetical order, as g, h,

Answer 2

Step-by-step explanation:

We have,

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

⇒

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

${x}^{2} - 2( \frac{ \sqrt{3 + 1} }{2} )x + ( \frac{ \sqrt{3 + 1} }{2} {)}^{2}$

$- \sqrt{3} + ( \frac{ \sqrt{3} + 1 }{2} {)}^{2}$

$(x - \frac{ \sqrt{3} + 1 }{2} {)}^{2} = \frac{ - 4 \sqrt{3} + (\sqrt{3} + 1{)}^{2} }{4}$

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

$x - \frac{ \sqrt{3} + 1}{2} = + \ - \frac{ \sqrt{3} - 1 }{2}$

$x = \frac{ \sqrt{3} + 1 }{2} + - \frac{ \sqrt{3} - 1}{2}$

$\: \: \: \: x = \sqrt{3} </em></strong><strong><em>,</em></strong><strong><em>\: 1$

˙❥ Hope it helps you.❤️