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CAT 2025 Lesson : Quadratic Equations - Infinite Series & Transposed

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4.4 Recurring Infinite Series

Example 16

If x=15−215−215−...∞x = \sqrt{15 - 2 \sqrt{15 - 2 \sqrt{15 - ... \infty}}} and x>0x \gt 0, what is the value of xx?

(1) 33            (2) −3-3            (3) 55            (4) −5-5           

Solution

x=15−215−215−...∞x = \sqrt{15 - 2 \sqrt{15 - 2 \sqrt{15 - ... \infty}}}
⇒ x=15−2x x = \sqrt{15 - 2x}
⇒x2=15−2xx^{2} = 15 - 2x
⇒ x2+2x−15=0x^{2} + 2x - 15 = 0
⇒ x=−5,3x = -5, 3 (−5(-5 is rejected as x>0)x \gt 0)

Answer: (1) 33


Example 17

1+121+121+121+...1 + \dfrac{12}{1 + \dfrac{12}{1 + \dfrac{12}{1 + ...}}}

(1) 44            (2) −3-3            (3) 44 or −3-3            (4) Cannot be determined           

Solution

Let 1+121+121+121+...1 + \dfrac{12}{1 + \dfrac{12}{1 + \dfrac{12}{1 + ...}}}

⇒ x=1+12x x = 1 + \dfrac{12}{x}    ⇒    x2=x+12 x^{2} = x + 12

⇒ x2−x−12=0x^{2} - x - 12 = 0

⇒ (x−4)(x+3)=0 (x - 4) (x + 3) = 0

⇒ x=4 x \bm{= 4} [x=−3x = -3 is rejected as there is no negative term in the initial expression]

Answer: (1) 4(1) \space 4


4.5 Interchanged/Transposed

Example 18

For how many integral values of xx is (x+12x−1)−10(2x−1x+1)=−3\left(\dfrac{x + 1}{2x - 1} \right) - 10 \left(\dfrac{2x - 1}{x + 1} \right) = -3?

Solution

Let y=x+12x−1y = \dfrac{x + 1}{2x - 1}

(x+12x−1)−10(2x−1x+1)=−3\left(\dfrac{x + 1}{2x - 1} \right) - 10 \left(\dfrac{2x - 1}{x + 1} \right) = -3    ⇒    y−10y=−3 y - \dfrac{10}{y} = -3

y2+3y−10=0y^{2} + 3y - 10 = 0    ⇒    (y+5)(y−2)=0 (y + 5) (y - 2) = 0    ⇒    y=2,−5 y = 2, -5

Substituting the value of yy,
If x+12x−1=2\dfrac{x + 1}{2x - 1} = 2    ⇒    x+1=4x−2 x + 1 = 4x - 2

⇒ x=1 x = 1 (Integral solution)

If x+12x−1=−5\dfrac{x + 1}{2x - 1} = -5    ⇒    x+1=−10x+5 x + 1 = -10x + 5

⇒ x=411 x = \dfrac{4}{11} (Non-integral solution)

∴ There is only 11 integral solution for xx.

Answer: 11


Example 19

How many distinct real values of xx satisfy 3(x2+1x2)−16(x+1x)+26=03 \left(x^{2} + \dfrac{1}{x^{2}} \right) - 16 \left(x + \dfrac{1}{x} \right) + 26 = 0?

Solution

(x+1x)2=x2+1x2+2\left(x + \dfrac{1}{x} \right)^{2} = x^{2} + \dfrac{1}{x^{2}} + 2    ⇒    x2+1x2=(x+1x)2−2 x^{2} + \dfrac{1}{x^{2}} = \left(x + \dfrac{1}{x} \right)^{2} - 2

Let (x+1x)=y\left(x + \dfrac{1}{x} \right) = y

3(x2+1x2)−16(x+1x)+26=03 \left(x^{2} + \dfrac{1}{x^{2}} \right) - 16 \left(x + \dfrac{1}{x} \right) + 26 = 0    ⇒    3(y2−2)−16y+26=0 3(y^{2} - 2) - 16y + 26 = 0

⇒ 3y2−16y+20=0 3y^{2} - 16y + 20 = 0

y=103,2y = \dfrac{10}{3}, 2

Substituting each of these values,
(x+1x)=103\left(x + \dfrac{1}{x} \right) = \dfrac{10}{3}    ⇒    3x2−10x+3=0 3x^{2} - 10x + 3 = 0    ⇒    x=3,13 x = \bm{3, \dfrac{1}{3}}

(x+1x)=2\left(x + \dfrac{1}{x} \right) = 2    ⇒    x2−2x+1=0 x^{2} - 2x + 1 = 0    ⇒    x=1 x = \bm{1}

Answer: 33 values


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