Detailed Notes: Quadratic Equations

Table of Contents
1. What are Quadratic Equations?
2. Factor and Remainder Theorem
3. Roots of an Equation
4. Methods of Solving Quadratic Equations
5. Algebraic Method
6. Solved Examples: Algebraic Method
7. Graphical Method
8. Solved Examples: Graphical  Method
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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

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

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.

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.

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

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