Question

$y = \sqrt{4 + \sqrt{4 + \sqrt{4 + ......} } }$
find y

Answer 1

$y=\sqrt{4+\sqrt{4+\sqrt{4+......}}}$

⇒  Square both the sides.

$y^2=\left(\sqrt{4+\sqrt{4+\sqrt{4+......}}}\right)^2 \\ \\ \\ y^2=4+\sqrt{4+\sqrt{4+......}}$

⇒  Subtract 4 from both sides.

$y^2-4=4+\sqrt{4+\sqrt{4+......}}-4 \\ \\ \\ y^2-4=\sqrt{4+\sqrt{4+......}}$

⇒  Now the RHS became its first form. Take RHS as y.

$y^2-4=y\ \ \ \ \ \ \ \ \ \ \left[\because y=\sqrt{4+\sqrt{4+......}}\right]$

⇒  Subtract y from both sides.

$y^2-y-4=0$

⇒  Now, factorize.

$y^2-y-4=0 \\ \\ \\ \\ \displaystyle y^2 \ - \ \frac{1-\sqrt{17}}{2}y \ - \ \frac{1+\sqrt{17}}{2}y \ - \ 4 \ =\ 0 \\ \\ \\ \\ y\left(y-\frac{1-\sqrt{17}}{2}\right)-\frac{1+\sqrt{17}}{2}\left(y-\frac{1-\sqrt{17}}{2}\right)=0 \\ \\ \\ \\ \left(y-\frac{1+\sqrt{17}}{2}\right)\left(y-\frac{1-\sqrt{17}}{2}\right)=0 \\ \\ \\ \\ \\ \large y=\text{ \bold{\frac{1+\sqrt{17}}{2}} }\ \ \ \ \ ; \ \ \ \ \ y=\text{ \bold{\frac{1-\sqrt{17}}{2}} }$

Thus we got the answer!

$\displaystyle \large y=\text{ \bold{\frac{1 \pm \sqrt{17}}{2}} }$

Answer 2

$\huge{\boxed{\mathfrak{Answer\::-}}}$

♦ As we are provided that

$y = \sqrt{4 + \sqrt{4 +\sqrt{4 + ...}}}$

♦ As it goes to the infinity

Then by Squaring both sides we get

$y^2 = \sqrt{4 + \sqrt{4 +\sqrt{4 + ...}}}^2$

$y^2 = 4 + \sqrt{4 + \sqrt{4 +\sqrt{4 + ...}}}$

♦ Now as $y = \sqrt{4 + \sqrt{4 +\sqrt{4 + ...}}}$ we can write above as

$y^2 = 4 + y$

♦ Now by making RHS = 0 .

$y^2 - y - 4 = 0$

• We get a quadratic equation

♦ Now by using quadratic formula that .

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

>> Here a,b,c are the numeric values

x = y

a = 1

b = -1

c = -4

♦ So by substituting values of a , b and c

$y = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)}$

$y = \dfrac{1 \pm \sqrt{1 + 16}}{2}$

$y = \dfrac{1 \pm \sqrt{17}}{2}$

♦ So value of y is either

$y = \dfrac{1 + \sqrt{17}}{2}$

or

$y = \dfrac{1 - \sqrt{17}}{2}$