Question

1/2+root3 + 1/2-root3​. plz answer

Answer 1

Answer:

1/2 + √3 + 1/2 - √3​ = 4

Step-by-step explanation:

Let

x = 1/2+√3

y = 1/2 - √3

• We have to find the value of (x + y)

Find the value of x :

Rationalizing factor = 2 - √3

Multiply and divide the fraction by (2 - √3)

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

Find the value of y :

Rationalizing factor = 2 + √3

Multiply and divide the fraction by (2 + √3)

 $\sf y=\dfrac{1}{2-\sqrt{3}} \times \dfrac{2+\sqrt{3}}{2+\sqrt{3}} \\\\\\ \sf y=\dfrac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} \\\\\\ \sf y=\dfrac{2+\sqrt{3}}{2(2+\sqrt{3})-\sqrt{3}(2+\sqrt{3})} \\\\\\ \sf y=\dfrac{2+\sqrt{3}}{4+2\sqrt{3}-2\sqrt{3}-\sqrt{3}^2} \\\\\\ \sf y=\dfrac{2+\sqrt{3}}{4-3} \\\\ \sf y=\dfrac{2+\sqrt{3}}{1} \\\\ \sf y=2+\sqrt{3}$

➣ x = 2 - √3

➣ y = 2 + √3

➟ x + y

➟ 2 - √3 + 2 + √3

➟ 2 + 2

➟ 4

Therefore, 1/2 + √3 + 1/2 - √3​ = 4

Answer 2

Step-by-step explanation:

$\boldsymbol {\huge{Solution:-}}$

$\implies \red{ \dfrac{1}{2 + \sqrt{3} } + \dfrac{1}{2 - \sqrt{3} } }$

Cross multiply and we know that,

(a+b)(a-b) = a²-b²

$\implies \green{ \dfrac{2 - \sqrt{3} + 2 + \sqrt{3} }{ {2}^{2} - {( \sqrt{3} )}^{2} } }$

$\implies \green { \dfrac{4}{4 - 3} }$

$\implies \green{4}$

$\boldsymbol {\huge{\pink{{ \therefore \: \dfrac{1}{2 + \sqrt{3} } + \dfrac{1}{2 - \sqrt{3} } } = 4}}}$

Make it easy !!!