model 5 simplifying roots of roots Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (d)

Let x = $√{6+√{6+√{6 +...}}}$

Squaring on both sides,

$x^2 = 6+√{6+√{6+√{6 +...}}}$

$x^2$ = 6 + x

$x^2$ - x - 6 = 0

$x^2$ - 3x + 2x - 6 = 0

x (x - 3) + 2 (x - 3) = 0

(x + 2) (x - 3) = 0

x = 3 because x ≠ - 2

$√{6+√{6+√{6 +...}}}$ = 3

It is because

6 = 2 × 3 = n (n + 1)

Answer: (a)

Let x = $√{1+√{1+√{1 +...}}}$

On squaring both sides

$x^2 =1+√{1+√{1+√{1 +...}}}$

$x^2$ = 1 + x

$x^2$ - x - 1 = 0

$x = {+1± √{1+4}}/2 = {+1 ±√ 5}/2$

But sum of + ve numbers can't be negative.

$x={1+ √5}/2 ={1+ 2.236}/2$

= ${3.236}/2 =1.618$

Thus 1 < 1.618 < 2

Answer: (b)

Let x = $√{12+√{12+√{12 +...}}}$

On squaring both sides,

$x^2 =12+√{12+√{12+√{12 +...}}}$

$x^2$ = 12 + x

$x^2$ - x - 12 = 0

$x^2$ - 4x + 3x - 12 = 0

x (x - 4) + 3 (x - 4) = 0

(x - 4) (x + 3) = 0

x = 4, - 3

The given expression is positive.

x = 4

$√{12+√{12+√{12}}}$=4

It is because

12 = 3 × 4 = n (n + 1)

Answer: (b)

Let x = $√{3√{3√{3...}}}$

Squaring both sides,

$x^2 = 3√{3√{3√{3...}}}$ = 3x

$x^2$ - 3x = 0

x (x - 3) = 0

x = 3 because x ≠ 0

$√{3+√{3+√{3+...∞}}}$ = 3

It is because, here

n = ∞ and x =3

$√{3+√{3+√{3+...∞}}}=3^(1-1/{3∞})$

= $3^(1 - 0)$ [${something}/∞ = 0$] = 3

Answer: (c)

m = $√{5+√{5+√{5 +...}}}$

On squaring both sides,

$m^2 = 5 + m ⇒ m^2$ - m = 5 ....(i)

Again,

n = $√{5-√{5-√{5-...}}}$

On squaring both sides,

$n^2$ = 5 - n

$n^2$ + n = 5 .........(ii)

$m^2$ - m = $n^2$ + n

$(m^2 - n^2)$ = m + n

(m + n) (m - n) - (m + n) = 0

(m + n) (m - n - 1) = 0

Answer: (d)

x = $√{72+√{72+√{72 +...}}}$

On squaring both sides,

$x^2 = 72+√{72+√{72+√{72 +...}}}$

$x^2$ = 72 + x

$x^2$ - x - 72 = 0

$x^2$ - 9x + 8x - 72 = 0

x (x - 9) + 8 (x - 9) = 0

(x + 8) (x - 9) = 0

x = 9 because x ≠ - 8

Using Rule 25

$√{72+√{72+√{72 +...}}}$ = 9

It is because 72 = 8×9 = n (n + 1)