Question

If a>0 then then root of a+ root of a + root of a+........infinity=......

Answer 1

✪ANSWER✪

• $\frac{ 1 + \sqrt{ 1 + 4a}}{2}$

★EXPLAINATION★

☆GIVEN

• $\sqrt{a + \sqrt{a + \sqrt{a +.....∞}}}$

• a > 0

✩ᴛᴏ ғɪɴᴅ

• Value of given expression

☆ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ

Let

$\sqrt{a + \sqrt{a + \sqrt{a +..∞}}}=y, y>0$

$⇒\sqrt{a + y}=y$

$⇒ a + y =y²$

$⇒ y² - y - a = 0$

$⇒ y = \frac{ 1 ± \sqrt{ 1² - 4\times 1\times (-a)}}{2\times 1}$

$⇒ y = \frac{ 1 ± \sqrt{ 1 + 4a}}{2}$

Since $y>0$

$⇒ y = \frac{ 1 + \sqrt{ 1 + 4a}}{2}$

Answer 2

Correct question: If a>0 then, $\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+.......}} } } =?$

Answer:

The value of the expression is $\frac{1+\sqrt{5} }{2}$.

Given,

An expression: $\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+.......}} } }$.

To Find,

The value of the expression.

Solution,

The method of finding the value of the expression is as follows -

Let $y=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+.......}} } }$.

Then, $y^2=a+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+.......}} } }$

⇒  $y^2=a+y$ [Since $y=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+.......}} } }$]

⇒  $y^2-y-a=0$

This is a quadratic equation. So we can solve it by the following method.

$y=\frac{-(-1)+\sqrt{(-1)^2-4*1*(-1)} }{2*1} =\frac{1+\sqrt{5} }{2}$ or, $y=\frac{-(-1)-\sqrt{(-1)^2-4*1*(-1)} }{2*1} =\frac{1-\sqrt{5} }{2}$.

But $\frac{1-\sqrt{5} }{2} < 0$, so $y\neq\frac{1-\sqrt{5} }{2}$.

So, $y=\frac{1+\sqrt{5} }{2}$.

Hence, the value of the expression is $\frac{1+\sqrt{5} }{2}$.

#SPJ2