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Quadratic Equations

Quadratic Equations

MODULES

Basics of Polynomial & Quadratic Equations
Discriminant & Graphical Representation
Sum & Product of Roots
Factorisation Method
Formulation & Completion of Squares
Changes to Roots
Mistakes in Roots, Common Roots & Squaring
Infinite Series & Transposed
Other Types
Higher Order Equations
Synthetic Division & Remainder Theorem
Maxima, Minima & Descrates Rule
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Quadratic Equations 1
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Quadratic Equations 2
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Quadratic Equations 3
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PRACTICE

Quadratic Equations : Level 1
Quadratic Equations : level 2
Quadratic Equations : level 3
ALL MODULES

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}}}x=15−215−215−...∞​​​ and x>0x \gt 0x>0, what is the value of xxx?

(1) 333            (2) −3-3−3            (3) 555            (4) −5-5−5           

Solution

x=15−215−215−...∞x = \sqrt{15 - 2 \sqrt{15 - 2 \sqrt{15 - ... \infty}}}x=15−215−215−...∞​​​
⇒ x=15−2x x = \sqrt{15 - 2x}x=15−2x​
⇒x2=15−2xx^{2} = 15 - 2xx2=15−2x
⇒ x2+2x−15=0x^{2} + 2x - 15 = 0x2+2x−15=0
⇒ x=−5,3x = -5, 3x=−5,3 (−5(-5(−5 is rejected as x>0)x \gt 0)x>0)

Answer: (1) 333


Example 17

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

(1) 444            (2) −3-3−3            (3) 444 or −3-3−3            (4) Cannot be determined           

Solution

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

⇒ x=1+12x x = 1 + \dfrac{12}{x}x=1+x12​    ⇒    x2=x+12 x^{2} = x + 12x2=x+12

⇒ x2−x−12=0x^{2} - x - 12 = 0x2−x−12=0

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

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

Answer: (1) 4(1) \space 4(1) 4


4.5 Interchanged/Transposed

Example 18

For how many integral values of xxx 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(2x−1x+1​)−10(x+12x−1​)=−3?

Solution

Let y=x+12x−1y = \dfrac{x + 1}{2x - 1}y=2x−1x+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(2x−1x+1​)−10(x+12x−1​)=−3    ⇒    y−10y=−3 y - \dfrac{10}{y} = -3y−y10​=−3

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

Substituting the value of yyy,
If x+12x−1=2\dfrac{x + 1}{2x - 1} = 22x−1x+1​=2    ⇒    x+1=4x−2 x + 1 = 4x - 2x+1=4x−2

⇒ x=1 x = 1x=1 (Integral solution)

If x+12x−1=−5\dfrac{x + 1}{2x - 1} = -52x−1x+1​=−5    ⇒    x+1=−10x+5 x + 1 = -10x + 5x+1=−10x+5

⇒ x=411 x = \dfrac{4}{11}x=114​ (Non-integral solution)

∴ There is only 111 integral solution for xxx.

Answer: 111


Example 19

How many distinct real values of xxx 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 = 03(x2+x21​)−16(x+x1​)+26=0?

Solution

(x+1x)2=x2+1x2+2\left(x + \dfrac{1}{x} \right)^{2} = x^{2} + \dfrac{1}{x^{2}} + 2(x+x1​)2=x2+x21​+2    ⇒    x2+1x2=(x+1x)2−2 x^{2} + \dfrac{1}{x^{2}} = \left(x + \dfrac{1}{x} \right)^{2} - 2x2+x21​=(x+x1​)2−2

Let (x+1x)=y\left(x + \dfrac{1}{x} \right) = y(x+x1​)=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 = 03(x2+x21​)−16(x+x1​)+26=0    ⇒    3(y2−2)−16y+26=0 3(y^{2} - 2) - 16y + 26 = 03(y2−2)−16y+26=0

⇒ 3y2−16y+20=0 3y^{2} - 16y + 20 = 03y2−16y+20=0

y=103,2y = \dfrac{10}{3}, 2y=310​,2

Substituting each of these values,
(x+1x)=103\left(x + \dfrac{1}{x} \right) = \dfrac{10}{3}(x+x1​)=310​    ⇒    3x2−10x+3=0 3x^{2} - 10x + 3 = 03x2−10x+3=0    ⇒    x=3,13 x = \bm{3, \dfrac{1}{3}}x=3,31​

(x+1x)=2\left(x + \dfrac{1}{x} \right) = 2(x+x1​)=2    ⇒    x2−2x+1=0 x^{2} - 2x + 1 = 0x2−2x+1=0    ⇒    x=1 x = \bm{1}x=1

Answer: 333 values


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