Quadratic Equation Notes for IIT JEE

Quadratic Equation Notes for IIT JEE contain all the important formulas, identities, theorems, properties, and equations. Students are recommended to learn Quadratic Equation Notes for IIT JEE. Revising these Quadratic Equation Notes for IIT JEE will help them easily memorise the important formulas and main topics.

We provide complete notes of Quadratic Equation for IIT JEE. Students can easily download the Quadratic Equation Notes for IIT JEE PDF for free from the link provided below.

Click here to download the PDF: – Download Quadratic Equation Notes for IIT JEE

Quadratic Equation

A quadratic equation is a second-degree equation. The general form of the quadratic equation is ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. For example, x² + 2x + 1 = 0.

A polynomial is an algebraic expression containing many terms. A polynomial of degree two of the form ax² + bx + c (a ≠ 0) is called a quadratic expression in x. We can also say that a quadratic polynomial f(x), when equated to zero, is called a quadratic equation.

Table of Contents

• Roots of Quadratic Equation
• Sum and Product of Roots
• Nature of Roots
• Graph of Quadratic Equation
• Quadratic Expression in Two Variables
• Solved Examples
• Practice Problems
• Frequently Asked Questions

Roots of Quadratic Equation

The values of x which satisfy the quadratic equation are called the roots of the quadratic equation. Roots are also called zeros or solutions of a quadratic equation.

Solving Quadratic Equation

a) Using the quadratic formula: If ax² + bx + c = 0 is the given quadratic equation, the roots are given by x = [-b ± √(b² – 4ac)]/2a.

b) Using the factorisation method: If ax² + bx + c = a(x – α)(x – β) = 0, then x = α and x = β will satisfy the given equation.

So, factorising the equation and equating each to zero gives the roots of the given equation.

Sum and Product of Roots

If α and β are the roots of the quadratic equation ax² + bx + c = 0, then

Sum of roots, α + β = -b/a

Product of roots, αβ = c/a

Discriminant of Quadratic Equation

The discriminant of the quadratic equation ax² + bx + c = 0 is given by D = b² – 4ac. It is also denoted by ∆.

Nature of Roots

If D = 0, the roots are real and equal.

If D > 0, the roots are real and unequal.

If D < 0, the roots are imaginary and unequal.

If D > 0 and D is a perfect square, the roots are rational and unequal.

If D > 0 and D is not a perfect square, the roots are irrational and unequal.

Formation of Quadratic Equation with Given Roots

If α and β are the roots of the quadratic equation, then

(x – α)(x – β) = 0

⇒ x² – (α + β)x + αβ = 0

⇒ x² – (sum of roots)x + product of roots = 0

Conjugate roots: The irrational and complex roots of a quadratic equation with a rational coefficient always occur in pairs. So, if one root is α + iβ, then the other root will be α – iβ.

Relation between Roots and Coefficients

If α and β are the roots of quadratic equation ax² + bx + c = 0, then

(a) α² + β² = (α + β)² – 2αβ = (b² – 2ac)/a²

(b) α – β = √((α + β)² – 4αβ) = ±√(b² – 4ac)/a = ±√D/a

(c) α² – β² = (α + β)√((α + β)² – 4αβ) = -b√(b² – 4ac)/a²

(d) α³ – β³ = (α – β)³ + 3αβ(α – β) = (b² – ac)√(b² – 4ac)/a³

(e) α³ + β³ = (α + β)³ – 3αβ(α + β) = -b(b² – 3ac)/a³

(f) α² + αβ + β² = (α + β)² – αβ

Roots under Special Cases

Consider the quadratic equation ax² + bx + c = 0

(i) If c = 0, then one root is zero, and another root is -b/a.

(ii) If b = 0, the roots are equal but opposite in sign.

(iii) If b = c = 0, then both roots are zero.

(iv) If a = c, then the roots are reciprocal to each other.

(v) If a + b + c = 0, then one root is 1, and the second root is c/a.

(vi) If a = b = c = 0, then the equation will become an identity and will satisfy every value of x.

Nature of Factors of the Quadratic Expression

The nature of factors of the quadratic expression ax² + bx + c is the same as the nature of roots of the corresponding quadratic equation ax² + bx + c = 0. Hence, the factors of the expression are as follows:

(i) Real and different, if b² – 4ac>0.

(ii) Rational and different, if b² – 4ac is a perfect square.

(iii) Real and equal, if b² – 4ac = 0.

(iv) Imaginary, if b² – 4ac < 0.

Condition for Common Roots

(a) Only one common root

Consider the following quadratic equations:

a₁x² + b₁x + c₁ = 0

a₂x² + b₂x + c₂ = 0

Let α be the common root of the above quadratic equations.

So, we can write

a₁α² + b₁α + c₁ = 0

a₂α² + b₂α + c₂ = 0

The condition for only one root common:

(c₁a₂ – c₂a₁)² = (b₁c₂ – b₂c₁)(a₁b₂ – a₂b₁).

(b) Both roots are common

The condition for both roots is common:

a₁/a₂ = b₁/b₂ = c₁/c_2.

Graph of Quadratic Equation

Consider a quadratic equation ax² + bx + c = 0, where a, b, and c are real and a ≠ 0.

The graph of a quadratic equation is a parabola.

If a>0, then the graph of a quadratic equation will be concave upwards.

If a<0, then the graph of a quadratic equation will be concave downwards.

The value of the discriminant determines whether the graph of a quadratic equation will:

• Intersect the x-axis at two points, i.e., b² – 4ac > 0
• Just touches the x-axis, i.e., b² – 4ac = 0
• Never intersects the x-axis, i.e., b² – 4ac < 0

Case 1: If a > 0, D > 0

The graph of a quadratic equation will be concave upwards and will intersect the x-axis at two points, α and β. The quadratic equation will have two real roots (α and β), and the curve will always lie above the x-axis.

Case 2: If a > 0, D = 0

The graph of a quadratic equation will be concave upwards and will intersect the x-axis at one point (-b/2a). The quadratic equation will have two equal roots (α = β).

Case 3: If a > 0, D < 0

The graph of a quadratic equation will be concave upwards and will not intersect the x-axis. The quadratic equation will have imaginary roots.

Case 4: If a < 0, D > 0

The parabola will be concave downwards and will intersect the x-axis at two points, α and β. The quadratic equation will have two real roots.

Case 5: If a < 0, D = 0

The parabola will be concave downwards and will intersect the x-axis at one point. The quadratic equation will have two equal roots.

Case 6: If a < 0, D < 0

The parabola will be concave downwards and will not intersect the x-axis. The quadratic equation will have imaginary roots.

Maximum and Minimum Value

Consider the quadratic expression ax² + bx + c.

a) If a < 0, then the expression has the greatest value at x = -b/2a. The maximum value is -D/4a.

b) If a > 0, then the expression has the least value at x = -b/2a. The minimum value is -D/4a.

Quadratic Expression in Two Variables

The general form of a quadratic equation in two variables, x and y, is

ax² + 2hxy + by² + 2gx + 2fy + c = 0.

To resolve the above expression into 2 linear rational factors, the condition is

∆ = 0

⇒ abc + 2fgh – af² – bg² – ch² = 0 and h² – ab > 0

This is called the discriminant of the given expression.

Important Points to Remember

(i) If α is a root of the equation f(x) = 0, then polynomial f(x) is exactly divisible by (x – α). This implies that (x – α) is a factor of f(x).

(ii) Every equation of n^th (n ≥ 1) degree has exactly n roots, and if the equation has more than n roots, then it is an identity.

(iii) If roots of quadratic equation a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 are in the same ratio,

i.e., α₁/β₁ = α₂/β_2, then b₁²/b₂² = a₁c₁/a₂c₂.

(iv) Quadratic equations having a modulus sign are solved considering both positive and negative values of the quantity containing the modulus sign. The roots of the given equation are those values among the value of the variable thus obtained, which satisfy the given equation.

(v) If one root is k times the other root of a quadratic equation a₁x² + b₂x + c₁ = 0, then

(k + 1)²/k = b²/ac.

Solved Examples

Example 1.

If α and β are the roots of the equation x² – x + 1 = 0, then α²⁰⁰⁹ + β²⁰⁰⁹ =

(a) -2

(b) -1

(c) 1

(d) 2

Solution:

Given x² – x + 1 = 0

Using quadratic formula, we get x = (1 ± i√3)/2

α = -ω, β = -ω²

α²⁰⁰⁹ +β²⁰⁰⁹ = -ω²⁰⁰⁹ + (-ω²)²⁰⁰⁹

= -(ω² + ω)

= 1

Hence, option c is the answer.

Example 2:

If p and q are the roots of the equation x² + px + q = 0, then

(a) p = 1, q = -2

(b) p = 0, q = 1

(c) p = -2, q = 0

(d) p = -2, q = 1

Solution:

Given x² + px + q = 0

Sum of roots, p + q = -p

Product of roots, pq = q

⇒ p = 1

1 + q = -1

⇒ q = -2

So p = 1, q = -2.

Hence, option a is the answer.

Example 3:

Let two numbers have an arithmetic mean 9 and a geometric mean 4. Then, these numbers are the roots of the quadratic equation

(a) x² + 18x + 16 = 0

(b) x² – 18x – 16 = 0

(c) x² + 18x – 16 = 0

(d) x² – 18x + 16 = 0

Solution:

Let m and n be the numbers.

(m + n)/2 = 9

(m + n) = 18

√(mn) = 4

mn = 16

So, the equation is x² – (sum of roots)x + product = 0

⇒ x² – 18x + 16 = 0

Hence, option d is the answer.

Example 4:

If one of the roots of the equation x² + px + 12 = 0 is 4, while the equation x² + px + q = 0 has equal roots, then the value of q is

(a) 49/4

(b) 4

(c) 3

(d) 12

Solution:

Given x² + px + 12 = 0 …(i)

Since 4 is a root of (i),

4² + 4p + 12 = 0

⇒ 4p = -28

⇒ p = -7

Given x² + px + q = 0 has equal roots.

So, D = 0

p² – 4q = 0

⇒ 49 – 4q = 0

⇒ q = 49/4

Hence, option a is the answer.

Example 5:

The value of a for which the sum of the squares of the roots of the equation x² – (a – 2)x – a – 1 = 0 assumes the least value

(a) 1

(b) 0

(c) 3

(d) 2

Solution:

Given x² – (a – 2)x – a – 1 = 0

Sum of roots, α + β = (a – 2)

Product of roots, αβ = -a – 1

Sum of squares of roots = α² + β² = (α + β)² – 2αβ

= a² – 2a + 6

= (a – 1)² + 5

⇒ a = 1

Hence, option a is the answer.

Practice Problems

1. The equation e^{sin x} – e^{-sin x} – 4 = 0 has

(a) No real roots

(b) Exactly one root

(c) Infinite number of real roots

(d) Exactly four real roots

2. If α and β are the distinct roots of the equation x² – x + 1 = 0, then α¹⁰¹ + β¹⁰⁷ is equal to

(a) 0

(b) 1

(c) -1

(d) 2

3. If the roots of the equation bx² + cx + a = 0 be imaginary, then for all real values of x, the expression 3b²x² + 6bcx + 2c² is

(a) greater than 4ab

(b) less than 4ab

(c) greater than -4ab

(d) less than -4ab

4. If x² + px – 444p = 0 has integral roots, where p is a prime number, then the value of p is

(a) 2

(b) 2, 3 and 37

(c) 3

(d) 37

5. If the roots of the equation x² – bx + c = 0 are two consecutive integers, then b²-4ac equals

(a) -2

(b) 3

(c) 2

(d) 1

Topmost Quadratic Equations Problems for JEE Advanced

Quadratic Equations – Flash Questions

Quadratic Equations – JEE Advanced Questions