Question

if y=2+√3 find the value
of a)
y + 1/2 b) y²+ 1 / y ^2​

Answer 1

a) 2 + √3 + 1/2 as usual

= 5/2 + √3

b)) given pic above

Answer 2

Given:

$y = 2 + \sqrt{3}$

To Find:

(a)

$y + \frac{1}{2}$

(b)

${y}^{2} + \frac{1}{ {y}^{2} }$

Solution:

For (a)

$y = 2 + \sqrt{3}$

$= y+ \frac{1}{2}$

adding the value of y in the equation formed,

$= (2 + \sqrt3) + \frac{1}{2}$

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

$= \frac{4 + 1}{2} + \sqrt{3}$

$= \frac{5}{2} + \sqrt{3}$

The value of $y + \frac{1}{2} = \frac{5}{2} + \sqrt{3}$

For (b)

${y}^{2} + \frac{1}{ {y}^{2} }$

where

$y = 2 + \sqrt{3}$

adding the value of y

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

changing the denominator to the numerator with a change in the sign.

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

using the formula of (a+b)² = a² + b² + 2ab and (a-b)² = a² + b² - 2ab

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

$= 4 + 3 + 4 \sqrt{3} + 4 + 3 - 4 \sqrt{3}$

$= 7 + 7$

$= 14$

Therefore, the value of ${y}^{2} + \frac{1}{ {y}^{2} } = 14$