Question

find the value of root 3 + root 5 upon 3 minus root 5 if root 5 equals to 2.236​

Answer 1

Answer:

$\sqrt{ \frac{(3 + \sqrt{5}) {}^{2} }{9 - 5} } \\ = \frac{3 + \sqrt{5} }{2} \\ = \frac{3 + 2.236}{2} \\ = \frac{5.236}{2}$

Answer 2

Answer:

$\qquad\boxed{\red{\sf \sqrt{\dfrac{(3+\sqrt5)}{(3-\sqrt5)}} = 2.618 }}$

Step-by-step explanation:

Given that the value of √5 is 2.236 . And we need to find out the value of given irrational number .The given number is ,

$\sf\dashrightarrow \sqrt{\dfrac{3+\sqrt5}{3-\sqrt5}}$

• Firstly we will rationalise the denominator by Multiplying the denominator with its conjugate irrational number . Since 3-√5 is present in the denominator , we will multiply it by 3+√5 .

$\sf\dashrightarrow \sqrt{\dfrac{(3+\sqrt5)(3+\sqrt5)}{(3-\sqrt5)(3+\sqrt5)}}$

• Now we can simply the denominator using the formula ,

$\sf\dashrightarrow\red{ a^2-b^2= (a+b)(a-b)}$

So that ,

$\sf\dashrightarrow \sqrt{\dfrac{(3+\sqrt5)(3+\sqrt5)}{(3)^2-(\sqrt5)^2}}$

$\sf\dashrightarrow \sqrt{\dfrac{(3+\sqrt5)^2}{9-5}}$

$\sf\dashrightarrow \sqrt{\dfrac{(3+\sqrt5)^2}{4}}$

• Now we can write 4 as 2² and the numerator has (3+√5)² . Hence on simplyfing the square root , we have ,

$\sf\dashrightarrow \sqrt{\dfrac{(3+\sqrt5)^2}{2^2}}$

$\sf\dashrightarrow \dfrac{3+\sqrt5}{2}$

• Now substituting the given value of √5 , we have ,

$\sf\dashrightarrow \dfrac{3+2.236}{2}$

$\sf\dashrightarrow \dfrac{5.236}{2}$

$\sf\dashrightarrow \boxed{\pink{\sf 2.618 }}$

Hence the required answer is 2.618 .

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$\tiny\qquad\qquad\qquad\qquad\boxed{\red{ \textsf{\textbf{ Hence the required answer is 2.618 .}}}}$

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