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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 - Formulation & Completion of Squares

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3.4.2 Formula Method

This method can be used when factorisation becomes difficult, i.e., you do not find roots by examining the sum and product of roots. This is especially the case when the roots are surds.

x=−b±D2a=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}x=2a−b±D​​=2a−b±b2−4ac​​

Therefore, if D=0D = 0D=0, then roots of the equation are real and equal, which is equal to −b2a\dfrac{-b}{2a}2a−b​; and if DDD is a square of a rational number, then the roots will be rational numbers.

Example 7

What are the roots of the equations x2+x−1=0x^{2} + x - 1 = 0x2+x−1=0 and 6x2−7x+2=06x^{2} - 7x + 2 = 06x2−7x+2=0?

Solution

x2+x−1=0x^{2} + x - 1 = 0x2+x−1=0 (a=1,b=1,c=−1)(a = 1, b = 1, c = -1)(a=1,b=1,c=−1)

⇒ x=−b±b2−4ac2a=−1±12−4×1×−12×1 x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \dfrac{-1 \pm \sqrt{1^{2} - 4 \times 1 \times -1}}{2 \times 1}x=2a−b±b2−4ac​​=2×1−1±12−4×1×−1​​ =−1±52= \dfrac{-1 \pm \sqrt{5}}{2}=2−1±5​​

⇒ x=−1+52,−1−52 x = \dfrac{-1 + \sqrt{5}}{2}, \dfrac{-1 - \sqrt{5}}{2}x=2−1+5​​,2−1−5​​

6x2−7x+2=06x^{2} - 7x + 2 = 06x2−7x+2=0 (a=6,b=−7,c=2)(a = 6, b = -7, c = 2)(a=6,b=−7,c=2)

⇒ x=−(−7)±49−482×6 x = \dfrac{-(-7) \pm \sqrt{49 - 48}}{2 \times 6}x=2×6−(−7)±49−48​​ =7±112=612,812=12,23= \dfrac{7 \pm 1}{12} = \dfrac{6}{12}, \dfrac{8}{12} = \dfrac{1}{2}, \dfrac{2}{3}=127±1​=126​,128​=21​,32​

Answer: 12,23\dfrac{1}{2}, \dfrac{2}{3}21​,32​


3.4.3 Completion of Squares

This can be used as an alternative to the Formula Method to enhance your speed. This should be used only when a is a perfect square and b is even, otherwise, finding the roots would become a tedious process.

Example 8

What are the roots of the equation 4x2−8x+3=04x^{2} - 8x + 3 = 04x2−8x+3=0 and 9x2+24x−13=09x^{2} + 24x - 13 = 09x2+24x−13=0?

Solution

Here, we express this quadratic equation as a2−2ab+b2a^{2} - 2ab + b^{2}a2−2ab+b2.

4x2−8x+3=04x^{2} - 8x + 3 = 04x2−8x+3=0
⇒ (2x)2−(2×2x×2)+22−1=0 (2x)^{2} - (2 \times 2x \times 2) + 2^{2} - 1 = 0(2x)2−(2×2x×2)+22−1=0
⇒ (2x−2)2=1 (2x - 2)^{2} = 1(2x−2)2=1     ⇒     2x−2=±1 2x - 2 = \pm 12x−2=±1     ⇒     2x=3,1 2x = 3, 1 2x=3,1     ⇒     3,1 3, 13,1     ⇒     x=32,12 x = \dfrac{3}{2}, \dfrac{1}{2}x=23​,21​

9x2+24x−13=09x^{2} + 24x - 13 = 09x2+24x−13=0
⇒ (3x)2+(2×3x×4)+42−29=0 (3x)^{2} + (2 \times 3x \times 4) + 4^{2} - 29 = 0(3x)2+(2×3x×4)+42−29=0
⇒ (3x+4)2=29(3x + 4)^{2} = 29(3x+4)2=29     ⇒     3x+4=±29 3x + 4 = \pm \sqrt{29}3x+4=±29​     ⇒     x=−4+293,−4−293 x = \dfrac{-4 + \sqrt{29}}{3}, \dfrac{-4 - \sqrt{29}}{3}x=3−4+29​​,3−4−29​​

Answer: −4+293,−4−293\dfrac{-4 + \sqrt{29}}{3}, \dfrac{-4 - \sqrt{29}}{3}3−4+29​​,3−4−29​​


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