IIT JEE Main Maths -Unit 2- Quadratic equations in real and complex systems- Study Notes-New Syllabus

IIT JEE Main Maths -Unit 2- Quadratic equations in real and complex systems – Study Notes – New syllabus

IIT JEE Main Maths -Unit 2- Quadratic equations in real and complex systems – Study Notes -IIT JEE Main Maths – per latest Syllabus.

Key Concepts:

Quadratic Equations in Real and Complex Systems
Solving Quadratic Equations by Different Methods
Newton’s Formula (Newton’s Identities / Newton’s Sums)

IIT JEE Main Maths -Study Notes – All Topics

Quadratic Equations in Real and Complex Systems

A quadratic equation is an equation of the form:

$ ax^2 + bx + c = 0 $, where $ a, b, c \in \mathbb{R} $ (or $ \mathbb{C} $), and $ a \ne 0 $.

The solutions (or roots) of the equation are the values of $ x $ that satisfy it.

1. Roots of a Quadratic Equation

The roots are given by the quadratic formula:

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

Here, the expression $ D = b^2 – 4ac $ is called the discriminant.

2. Nature of Roots (Real System)

Discriminant $D$	Nature of Roots	Remarks
$ D > 0 $	Real and distinct	Two different real roots
$ D = 0 $	Real and equal	Both roots are the same
$ D < 0 $	Complex conjugate	Non-real roots

3. Roots in Complex System

When $ D < 0 $, roots are complex. Let $ D = -k $ where $ k > 0 $.

$ x = \dfrac{-b \pm i\sqrt{k}}{2a} $

The roots are conjugate pairs of the form $ p + iq $ and $ p – iq $.

4. Relationship Between Coefficients and Roots

If $ \alpha $ and $ \beta $ are the roots of $ ax^2 + bx + c = 0 $, then:

Sum of roots: $ \alpha + \beta = -\dfrac{b}{a} $
Product of roots: $ \alpha \beta = \dfrac{c}{a} $

Example

Find the roots of $ 2x^2 – 3x + 1 = 0 $.

▶️ Answer / Explanation

Step 1: Identify coefficients $ a = 2, b = -3, c = 1 $.

Step 2: $ D = (-3)^2 – 4(2)(1) = 9 – 8 = 1 $

Step 3: $ x = \dfrac{3 \pm \sqrt{1}}{4} $

$ \Rightarrow x = 1 $ or $ x = \dfrac{1}{2} $

Answer: Roots are $ 1 $ and $ \dfrac{1}{2} $.

Example

Find the roots of $ x^2 + 4x + 8 = 0 $.

▶️ Answer / Explanation

Step 1: $ a = 1, b = 4, c = 8 $

Step 2: $ D = 4^2 – 4(1)(8) = 16 – 32 = -16 $

Step 3: $ x = \dfrac{-4 \pm i\sqrt{16}}{2} = \dfrac{-4 \pm 4i}{2} $

$ \Rightarrow x = -2 + 2i, \; x = -2 – 2i $

Answer: Complex conjugate roots are $ -2 \pm 2i $.

Example

Find the roots of $ 3x^2 + 2x + 5 = 0 $ and express in the form $ p \pm iq $.

▶️ Answer / Explanation

Step 1: $ a = 3, b = 2, c = 5 $

Step 2: $ D = b^2 – 4ac = 4 – 60 = -56 $

Step 3: $ x = \dfrac{-2 \pm i\sqrt{56}}{6} = \dfrac{-2 \pm i2\sqrt{14}}{6} $

$ \Rightarrow x = \dfrac{-1}{3} \pm i\dfrac{\sqrt{14}}{3} $

Answer: $ x_1 = -\dfrac{1}{3} + i\dfrac{\sqrt{14}}{3} $, $ x_2 = -\dfrac{1}{3} – i\dfrac{\sqrt{14}}{3} $.

Solving Quadratic Equations by Different Methods

A quadratic equation in one variable is of the form:

$ ax^2 + bx + c = 0, \quad a \ne 0 $

Its roots (solutions) are the values of $ x $ that satisfy this equation.

General Formula (Most Common Method)

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

where $ D = b^2 – 4ac $ is called the discriminant.

If $ D > 0 $ → Two distinct real roots
If $ D = 0 $ → Equal (repeated) real roots
If $ D < 0 $ → Complex conjugate roots

Method 1 — Factorization (Splitting the Middle Term)

This is the fastest and most JEE-friendly method when coefficients are rational.

Steps:

Multiply $ a \times c $.
Find two numbers whose product = $ a \times c $ and sum = $ b $.
Split the middle term accordingly and factorize.

Example

Solve $ 2x^2 + 5x + 3 = 0 $.

▶️ Answer / Explanation

Here, $ a \times c = 6 $ and $ b = 5 $.

Split $ 5x $ as $ 2x + 3x $: $ 2x^2 + 2x + 3x + 3 = 0 $.

$ 2x(x + 1) + 3(x + 1) = 0 \Rightarrow (2x + 3)(x + 1) = 0. $

Roots: $ x = -\dfrac{3}{2}, -1. $

Method 2 — Completing the Square

Useful when the equation is not easily factorable.

Steps:

Divide through by $ a $ to make the coefficient of $ x^2 $ unity.
Rearrange to get $ x^2 + \dfrac{b}{a}x = -\dfrac{c}{a} $.
Add and subtract $ \left(\dfrac{b}{2a}\right)^2 $ to complete the square.
Take square roots and simplify.

Example

Solve $ x^2 + 6x – 16 = 0 $.

▶️ Answer / Explanation

Move constant: $ x^2 + 6x = 16 $.

Add $ (6/2)^2 = 9 $: $ x^2 + 6x + 9 = 25. $

$ (x + 3)^2 = 25 \Rightarrow x + 3 = \pm 5. $

Roots: $ x = 2, -8. $

Method 3 — Using Vieta’s Relations (Sum and Product of Roots)

For $ ax^2 + bx + c = 0 $:

$ \alpha + \beta = -\dfrac{b}{a}, \quad \alpha\beta = \dfrac{c}{a} $

This trick helps to quickly form or find equations when roots are known or related.

Example

If one root is twice the other, and the equation is $ x^2 – 5x + 6 = 0 $, verify the condition.

▶️ Answer / Explanation

Let roots be $ \alpha $ and $ 2\alpha $.

Then, $ \alpha + 2\alpha = 3\alpha = 5 \Rightarrow \alpha = \dfrac{5}{3}. $

Product: $ 2\alpha^2 = 6 \Rightarrow \alpha^2 = 3 \Rightarrow \alpha = \sqrt{3}. $

Contradiction — means roots are not in 1:2 ratio. This trick helps check quickly without solving.

Method 4 — Graphical Method (Fast Intuition)

The quadratic $ y = ax^2 + bx + c $ represents a parabola:

If $ a > 0 $: opens upward.
If $ a < 0 $: opens downward.

The roots are the x-intercepts (where $ y = 0 $).

Trick: If $ D = b^2 – 4ac $ is negative → No x-intercept (complex roots).

Method 5 — Shortcut Using Sum and Product

If you know $ \alpha + \beta = S $ and $ \alpha\beta = P $, then roots are given directly by:

$ x = \dfrac{S \pm \sqrt{S^2 – 4P}}{2} $

This is faster than using $ a,b,c $ when only the relations are known.

Method 6 — Trick for Quadratics of Form $ (x + \dfrac{1}{x}) $

Many JEE problems involve transformations like:

$ x^2 + \dfrac{1}{x^2}, \quad x + \dfrac{1}{x} $

Trick: If $ t = x + \dfrac{1}{x} $, then

$ x^2 + \dfrac{1}{x^2} = t^2 – 2, \quad x^3 + \dfrac{1}{x^3} = t^3 – 3t $

Example

Solve $ x^2 + \dfrac{1}{x^2} = 3 $.

▶️ Answer / Explanation

Let $ t = x + \dfrac{1}{x} \Rightarrow x^2 + \dfrac{1}{x^2} = t^2 – 2. $

So $ t^2 – 2 = 3 \Rightarrow t^2 = 5 \Rightarrow t = \pm\sqrt{5}. $

Now $ x + \dfrac{1}{x} = \sqrt{5} \Rightarrow x^2 – \sqrt{5}x + 1 = 0. $

Roots: $ x = \dfrac{\sqrt{5} \pm 1}{2}. $

Method 7 — Substitution Trick (JEE Shortcut)

For equations like $ ax^4 + bx^2 + c = 0 $: Let $ x^2 = t \Rightarrow $ becomes a quadratic in $ t $.

Example

Solve $ x^4 – 5x^2 + 4 = 0. $

▶️ Answer / Explanation

Let $ x^2 = t \Rightarrow t^2 – 5t + 4 = 0. $

$ (t – 4)(t – 1) = 0 \Rightarrow t = 4, 1. $

$ \Rightarrow x = \pm2, \pm1. $

Roots: $ -2, -1, 1, 2. $

Method 8 — Observation & Coefficient Trick (JEE Shortcut)

When $ a + b + c = 0 $ → one root is $ x = 1 $. When $ a – b + c = 0 $ → one root is $ x = -1 $.

Example

Solve $ 3x^2 – 8x + 5 = 0. $

▶️ Answer / Explanation

Check: $ a + b + c = 3 – 8 + 5 = 0 \Rightarrow x = 1 $ is a root.

Divide by $ (x – 1) $: $ 3x^2 – 8x + 5 = (x – 1)(3x – 5). $

Roots: $ x = 1, \dfrac{5}{3}. $

Newton’s Formula (Newton’s Identities / Newton’s Sums)

Newton’s Formula provides a way to express the sum of powers of roots of a polynomial in terms of its coefficients.

It is especially useful for evaluating expressions like $ \alpha^2 + \beta^2 $, $ \alpha^3 + \beta^3 $, etc., without solving for the roots individually.

General Polynomial Form

Let

$ a_0x^n + a_1x^{n-1} + a_2x^{n-2} + \dots + a_{n-1}x + a_n = 0 $

with roots $ \alpha_1, \alpha_2, \dots, \alpha_n $.

Define the sum of k-th powers of roots as:

$ S_k = \alpha_1^k + \alpha_2^k + \dots + \alpha_n^k $

Newton’s General Formula

For all $ r \ge 1 $:

$ a_0S_r + a_1S_{r-1} + a_2S_{r-2} + \dots + a_{r-1}S_1 + r\,a_r = 0 $

This relation allows us to find $ S_r $ recursively using previous sums $ S_1, S_2, \dots, S_{r-1} $.

For Quadratic Equation

For $ ax^2 + bx + c = 0 $ with roots $ \alpha, \beta $:

$ \alpha + \beta = -\dfrac{b}{a} $
$ \alpha\beta = \dfrac{c}{a} $

Using Newton’s Sums:

$ S_1 = \alpha + \beta = -\dfrac{b}{a} $
$ S_2 = \alpha^2 + \beta^2 = S_1^2 – 2\dfrac{c}{a} $
$ S_3 = \alpha^3 + \beta^3 = S_1S_2 – S_1\dfrac{c}{a} $

For Cubic Equation

For $ ax^3 + bx^2 + cx + d = 0 $ with roots $ \alpha, \beta, \gamma $:

$ \alpha + \beta + \gamma = -\dfrac{b}{a} $
$ \alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{c}{a} $
$ \alpha\beta\gamma = -\dfrac{d}{a} $

Newton’s Sums:

$ S_1 = -\dfrac{b}{a} $
$ S_2 = \dfrac{b^2 – 2ac}{a^2} $
$ S_3 = -\dfrac{b^3 – 3abc + 3a^2d}{a^3} $

General Recursive Pattern

Newton’s relation for the first few terms:

r	Newton’s Identity
1	$ a_0S_1 + a_1 = 0 $
2	$ a_0S_2 + a_1S_1 + 2a_2 = 0 $
3	$ a_0S_3 + a_1S_2 + a_2S_1 + 3a_3 = 0 $
4	$ a_0S_4 + a_1S_3 + a_2S_2 + a_3S_1 + 4a_4 = 0 $

Shortcut Relations for Quadratics (JEE Tricks)

$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta $
$ \alpha^3 + \beta^3 = (\alpha + \beta)^3 – 3\alpha\beta(\alpha + \beta) $
$ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 – 2(\alpha\beta)^2 $

These can be directly derived from Newton’s identities and are faster for JEE-style questions.

Example

For $ x^2 – 5x + 6 = 0 $, find $ \alpha^2 + \beta^2 $.

▶️ Answer / Explanation

$ \alpha + \beta = 5, \alpha\beta = 6. $

$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta = 25 – 12 = 13. $

Answer: $ 13. $

Example

For $ x^3 – 6x^2 + 11x – 6 = 0 $, find $ \alpha^3 + \beta^3 + \gamma^3 $.

▶️ Answer / Explanation

$ a_0 = 1, a_1 = -6, a_2 = 11, a_3 = -6. $

$ S_1 = -\dfrac{a_1}{a_0} = 6, \quad S_2 = \dfrac{a_1^2 – 2a_0a_2}{a_0^2} = 36 – 22 = 14. $

Now use Newton’s formula for $ S_3 $: $ a_0S_3 + a_1S_2 + a_2S_1 + 3a_3 = 0. $

$ 1(S_3) – 6(14) + 11(6) – 18 = 0 \Rightarrow S_3 = 36. $

Answer: $ \alpha^3 + \beta^3 + \gamma^3 = 36. $

Example

For $ x^3 + 3x^2 + 3x + 1 = 0 $, find $ \alpha^5 + \beta^5 + \gamma^5 $.

▶️ Answer / Explanation

Given: $ a_0 = 1, a_1 = 3, a_2 = 3, a_3 = 1. $

$ S_1 = -3, \quad S_2 = (-3)^2 – 2(3) = 3, \quad S_3 = -3(3) – 3(-3) – 3(1) = -3. $

Now $ a_0S_4 + a_1S_3 + a_2S_2 + a_3S_1 + 4a_4 = 0. $

Since degree = 3, $ a_4 = 0 $. So $ S_4 + 3(-3) + 3(3) + 1(-3) = 0 \Rightarrow S_4 – 3 = 0 \Rightarrow S_4 = 3. $

Similarly, $ S_5 + 3S_4 + 3S_3 + S_2 = 0 \Rightarrow S_5 + 9 – 9 + 3 = 0 \Rightarrow S_5 = -3. $

Answer: $ \alpha^5 + \beta^5 + \gamma^5 = -3. $

Example

Let $ \alpha $ and $ \beta $ be the roots of the equation $ x^2 – 6x – 2 = 0 $. If $ a_n = \alpha^n – \beta^n $, for $ n \ge 1 $, then find the value of

$ \dfrac{a_{10} – 2a_8}{2a_9}. $

▶️ Answer / Explanation

Step 1: Given quadratic equation:

$ x^2 – 6x – 2 = 0. $

Hence, $ \alpha + \beta = 6 $ and $ \alpha\beta = -2. $

$ a_n = \alpha^n – \beta^n. $

Take common $ \alpha^8 $ in numerator.

$ a_{10} – 2a_8 = (\alpha^{10} – \beta^{10}) – 2(\alpha^8 – \beta^8) = \alpha^8(\alpha^2 – 2) – \beta^8(\beta^2 – 2). $

From the quadratic equation, $ \alpha^2 = 6\alpha + 2 $ and $ \beta^2 = 6\beta + 2. $

Therefore,

$ \alpha^2 – 2 = 6\alpha, \quad \beta^2 – 2 = 6\beta. $

Substitute these in the numerator:

$ a_{10} – 2a_8 = \alpha^8(6\alpha) – \beta^8(6\beta) = 6(\alpha^9 – \beta^9) = 6a_9. $

$ \dfrac{a_{10} – 2a_8}{2a_9} = \dfrac{6a_9}{2a_9} = 3. $

Notes and Study Materials

Examples and Exercise

IIT JEE (Main) Mathematics ,”Quadratic Equations” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian

About this unit

Quadratic equations in real and complex number system and their solutions. The relation between roots and coefficients, nature of roots, the formation of quadratic equations with given roots.

IITian Academy Notes for IIT JEE (Main) Mathematics – Quadratic Equations

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r	Newton’s Identity
1	\( a_0S_1 + a_1 = 0 \)
2	\( a_0S_2 + a_1S_1 + 2a_2 = 0 \)
3	\( a_0S_3 + a_1S_2 + a_2S_1 + 3a_3 = 0 \)
4	\( a_0S_4 + a_1S_3 + a_2S_2 + a_3S_1 + 4a_4 = 0 \)

IIT JEE Main Maths -Unit 2- Quadratic equations in real and complex systems- Study Notes-New Syllabus