Detailed Notes: Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced PDF Download

What are Quadratic Equations?

• A quadratic polynomial expression equated to zero becomes a quadratic equation 
• The values of x that satisfy the equation are called roots/ zeros of the Quadratic Equation. 
• General form: ax² + bx + c = 0, where a, b, c are real numbers, and a ≠ 0, the numbers a, b, and c are called the coefficients of the equation.

• a is called the leading coefficient, b is called the middle coefficient, and c is called the constant term.

• Examples: 3x² + x + 5 = 0,-x² + 7x + 5 = 0, x² + x = 0, x² = 0.

Factor and Remainder Theorem

Remainder Theorem

• The Remainder theorem states that when a polynomial p(x) is divided by a linear polynomial (x - a), then the remainder is equal to p(a)
• Consider f(x) = (x - r)q(x) + R.
• If you divide a polynomial f(x) by (x - c), the remainder of that division is equal to f(c).

Example: Use the remainder theorem to find the remainder when f(x) = 3x² + 5x - 8 is divided by (x - 2)

Sol: Use the Remainder Theorem. 
Put x = 2 in f(x). Since we are dividing f(x) = 3x² + 5x - 8 by (x - 2), we consider x = 2.
Hence, the remainder R is given by:
R = f(2) = 3(2²) + 5(2) - 8 = 14

Clear all your JEE doubts with EduRev Join for Free

Join for Free

Factor Theorem

• The Factor Theorem is a special case of the Remainder Theorem and is used to determine whether a given polynomial has a specific factor.
• If f(x) is a polynomial and (x - c) is a factor of f(x), then f(c) = 0.
• In other words:
  If you substitute x = c into the polynomial f(x) and the result is 0, then (x - c) is a factor of f(x).

Example: Use the factor theorem to decide if (x - 2) is a factor of f(x) = x⁵ - 2x⁴ + 3x³ - 6x² - 4x + 8. 

Sol: We know that (x - r) will be a factor of f(x) if f(r) = 0. Therefore, by using this condition, we can decide whether (x - 2) is a factor of the given polynomial or not.

f(x) = x⁵ - 2x⁴ + 3x³ - 6x² - 4x + 8. 
f(2) = (2)⁵ - 2(2)⁴ + 3(2)³ - 6(2)² - 4(2) + 8 = 0

Since f(2) = 0, we can conclude that (x - 2) is a factor.

Roots of an Equation

• The values of variables satisfying the given equation are called its roots
• In other words, x = α is a root of the equation f(x), if f(α) = 0. The real roots of an equation f(x) = 0 are the x-coordinates of the points where the curve y = f(x) intersects the x-axis.
• Example: x² - 3x + 2 = 0. At x = 1 and 2, the equation becomes zero.
• When referring to the zeroes of ax²+ bx + c, it usually means the roots of the equation ax²+ bx + c
  

Did You Know
A polynomial can be rewritten as given below: 
y = a(x - r₁)(x - r₂)....(x - r_n)
The factors like (x - r₁) are called linear factors because they describe a line when you plot them.

Example: What values of x satisfy the equation 2x² = 18?

Sol: 2x² = 18
2x² / 2 = 18 / 2
x² = 9
√(x²) = √9
x = ±3
The following values of x satisfy the equation 2x² = 18: 
-3 and 3

Methods of Solving Quadratic Equations

There are two methods to solve a Quadratic equation 

1. Algebraic Method
2. Graphical Method (Absolute)

Now we will discuss each method in detail to analyze the method to calculate roots

Algebraic Method

• As we have already done the middle-term splittin method but not every quadratic equation can be splitter into its respective factors
• Therefore we will use the Discriminant method prominently to solve equations
• The roots of a quadratic equation ax²+ bx + c = 0 are given by the quadratic formula:

  x = (-b ± √(b² - 4ac)) / 2a

• The term under the square root, b² - 4ac, is called the discriminant (Δ)

Did You Know

Let If α and β are the roots of a quadratic equation ax²+ bx + c and we have to calculate the sum and product of roots then

Sum of roots = - b/a

Product of roots = c/a 

Nature of Roots

1. D > 0: The roots are real and distinct (unequal).
2. $\mathrm{D\; =\; 0}$: The roots are real and coincident (equal).
3. D < 0: The roots are complex (imaginary).
4. If one root is $p+iq$, the other is its conjugate $p-iq$(where $p,q\in R$ and $i=-1$).

Very Important Conditions

• If y = ax² + bx + c is positive for all real values of x, then a > 0 and D < 0.
• If y = ax² + bx + c is negative for all real values of x, then a < 0 and D < 0.
• If both roots are infinite for the equation ax² + bx + c = 0,
  x = 1 / y ⇒ (a / y²) + (b / y) + c = 0.
• cy² + by + a = 0
  (- b / c) = 0, (a / c) = 0 ⇒ a = 0, b = 0, and c ≠ 0.

• The equations  a₁x² + b₁x + c₁ and   a₂x² + b₂x + c₂  have the following conditions:
  1. One common root if:
  (b₁c₂ - b₂c₁) / (c₁a₂ - c₂a₁) = (c₁a₂ - c₂a₁) / (a₁b₂ - a₂b₁)
  2. Both roots common if:
  a₁/a₂ = b₁/b₂ = c₁/c₂

Download the notes Detailed Notes: Quadratic Equations

Download as PDF

Download as PDF

Roots in Particular Cases

For the quadratic equation ax² + bx + c = 0:

(a) If b = 0, ac < 0 ⇒ Roots are of equal magnitude but of opposite sign
(b) If c = 0 ⇒ One root is zero, the other is -b/a;
(c) If b = c = 0 ⇒ Both roots are zero;
(d) If a = c ⇒ The roots are reciprocal to each other;
(e) If (a > 0; c < 0) or (a < 0; c > 0) ⇒ The roots are of opposite signs;
(f) If the sign of a = sign of b × sign of c ⇒ The root of greater magnitude is negative;
(g) If a + b + c = 0 ⇒ One root is 1 and the other is c/a;
(h) If a = b = c = 0, then the equation will become an identity and will be satisfied by every value of x.

Did You Know
If you have to find the maximum and minimum value of a quadratic equation ax² + bx + c

Write the quadratic equation ax² + bx + c into a * [(x + b / 2a)² - D / 4a²] , where  D = b² - 4ac

If a > 0:

• The equation has a minimum value of (4ac - b²) / 4a at x = -b / 2a.
• There is no maximum value.

If a < 0:

• The equation has a maximum value of (4ac - b²) / 4a at x = -b / 2a.
• There is no minimum value.

Solved Examples: Algebraic Method

Example 1:  Form a quadratic equation with real coefficients when one root is 3 - 2i. 

Sol: Since the complex roots always occur in pairs, so the other root is 3 + 2i. 
Therefore, we can form the required quadratic equation by obtaining the sum and product of the roots.

The sum of the roots is:
(3 + 2i) + (3 - 2i) = 6.

The product of the roots is:
(3 + 2i) × (3 - 2i) = 9 - 4i² = 9 + 4 = 13.

Hence, the equation is x² - Sx + P = 0.
⇒ x² - 6x + 13 = 0.

Example 2: Consider the quadratic polynomial f(x) = x² - px + q where f(x) = 0 has prime roots. If p + q = 11 and a = p² + q², then find the value of f(a) where a is an odd positive integer. 

Sol: Here f(x) = x² - px + q, hence by considering α and β as its roots and using the formulae for sum and product of roots and the given conditions, we get the values of f(a).

f(x) = x² - px + q
Given α and β are prime:
α + β = p ...(i)
αβ = q ...(ii)

Given p + q = 11 ⇒ α + β + αβ = 11
⇒ (α + 1)(β + 1) = 12; α = 2, β = 3 are the only primes that solve this equation.

∴ f(x) = (x - 2)(x - 3) = x² - 5x + 6
∴ p = 5, q = 6 ⇒ a = p² + q² = 25 + 36 = 51; 
f(51) = (51 - 2)(51 - 3) = 49 × 48 = 3422

Example 3: If a Quadratic equation (QE) is formed from y² = 4ax and y = mx + c and has equal roots, then find the relation between c, a, and m. 

Sol: By solving these two equations, we get the quadratic equation; and as it has equal roots, hence D = 0.

(mx + c)² = 4ax;
m²x² + 2(cm - 2a)x + c² = 0
Given that the roots are equal. So, D = 0 ⇒ 4(cm - 2a)² ⇒ 4c²m²
⇒ 4a² = 4acm
a = cm ⇒ c = a/m
This is a condition for the line y = mx + c to be a tangent to the curve y² = 4ax.

Take a Practice Test Test yourself on topics from JEE exam

Practice Now

Practice Now

Graphical Method

As we are talking about Quadratic Equations, 

• The graph of a quadratic equation ax²+ bx + c = 0 will always be a Parabola with vertex (-b/2a, -D/2a)
• Graph Shape:
  a > 0: Parabola opens upwards (concave up).
  a < 0: Parabola opens downwards (concave down).
• Graph Categories:

  1. Two real roots a and b where a < b:
  ax² + bx + c > 0 for x ∈ (-∞, a) ∪ (b, ∞),
  ax² + bx + c < 0 for x ∈ (a, b).
  2. One real root a = b:
  ax² + bx + c = 0 for x = a.
  3. No real roots:
  ax² + bx + c > 0 for x ∈ R (a > 0),
  ax² + bx + c < 0 for x ∈ R (a < 0).

• Some Important Cases are given below,
• 

Solved Examples: Graphical  Method

(A) b - c
(B) bc
(C) c - a
(D) ab²

Sol: Some observations from the graph are

• Direction of the parabola:

  • If it opens upward → a > 0.
  • If it opens downward → a < 0.

• Y-intercept:

  • The point where the graph crosses the y-axis is (0, c).
  • If the intercept is above the origin → c > 0.
  • If it is below the origin → c < 0.

• Vertex position:

  • The vertex is at (-b/2a, constant term).
  • If the vertex is left of the y-axis → -b/2a < 0.
  • If the vertex is right of the y-axis → -b/2a > 0.

• The slope at the y-intercept:

  • The derivative dy/dx = 2ax + b.
  • At x = 0, the slope is b. If the graph is rising at x = 0 → b > 0. If falling → b < 0.
• Here a < 0
• (-b/a) < 0 ⇒ b < 0; c > 0. 
• As b - c = (-ve) - (+ve), it must be negative;
• Also, bc = (-ve)(+ve); this must be negative;
• Then, β + (1/α) = (-ve) + (+ve); the product must be negative; finally,
• c - a = (+ve) - (-ve), it must be positive.

Example 2: Suppose the graph of a quadratic polynomial y = x² + px + q is situated so that it has two arcs lying between the rays y = x and y = 2x, x ≥ 0. These two arcs are projected onto the x-axis yielding segments S_L and S_R, with S_R to the right of S_L. Find the difference of the length (S_R) - (S_L). 

Sol: Let the roots of x² + px + q = x be x₁ and x₂, and the roots of x² + px + q = 2x be x₃ and x₄.

S_R = x₄ - x₂ and S_L = x₁ - x₃
⇒ S_R - S_L = x₄ + x₃ - x₁ - x₂.

∴ |(S_R) - (S_L)| = |[-(p - 2) - {-(p - 1)}]| = 1.

$a\; =\; b\; =\; c\; =\; 0$

$x^2\; -\; \backslash left(\; -\backslash frac\{b\}\{a\}\; \backslash right)x\; +\; \backslash frac\{c\}\{a\}\; =\; 0$