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Roots

Factorizing a quadratic expression

The primary goal of factorizing a quadratic expression \( ax^2 + bx + c \) is to simplify the equation and find its solutions more easily. Factorization transforms the quadratic into a product of binomials, typically in the form \( a(x - \alpha)(x - \beta) \), where \( \alpha \) and \( \beta \) are the roots.

🧠 Exploring the Factorization of Quadratic Expressions Using Completing the Square

One effective method to factorize a quadratic expression of the form \( ax^2 + bx + c \) is by completing the square. This method involves transforming the expression into a perfect square plus or minus a constant. Let's go through this process step-by-step:

➡ Step 1: Take common the coefficient of \(x^2\)

First, let's factor out the coefficient \( a \) from all terms in the expression:

\( a(x^2 + \frac{b}{a}x + \frac{c}{a}) \)

➡ Step 2: Completing the Square

Now, within the brackets, we add and subtract the square of half the coefficient of \( x \), which is \( \left(\frac{b}{2a}\right)^2 \):

\( a \left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} \right) \)

➡ Step 3: Forming a Perfect Square

We can rearrange the terms inside the brackets to form a perfect square:

\( a \left( \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} \right) \)

➡ Step 4: A rearrangement

Let \( \Delta = b^2 - 4ac \). Combine the last two terms:

\( a \left( \left(x + \frac{b}{2a}\right)^2 - \frac{{b^2-4ac}}{4a^2} \right) \)

\( = a \left( \left(x + \frac{b}{2a}\right)^2 - \frac{{\Delta}}{4a^2} \right) \)

➡ Step 5: Cases Based on \( \Delta \): 🔍
Whether the obtained expression can be factorized or not depends on \(\Delta\).

🔘 Case I. \( \Delta > 0 \):

  • In this case, \( \frac{\Delta}{4a^2} \) is a positive number.
  • The expression can be viewed as \(a^2-b^2=(a-b)(a+b)\). Factorization is possible: \( a\left(\left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2}\right) \) \(= a\left(\left(x + \frac{b}{2a}\right) - \frac{\sqrt{\Delta}}{2a}\right)\left(\left(x + \frac{b}{2a}\right) + \frac{\sqrt{\Delta}}{2a}\right) \)

🔘 Case II. \( \Delta = 0 \):

  • The expression becomes a perfect square: \( a\left(\left(x + \frac{b}{2a}\right)^2\right) \)
  • Factorization results in a single repeated factor.

🔘 Case III. \( \Delta < 0 \):🚫

  • The term \( \frac{\Delta}{4a^2} \) is negative, leading to \( a^2 + b^2 \) form inside the parentheses, which cannot be factorized.
  • Factorization is not possible into real factors. Even though we will learn later that we can factorize it into complex factors.

In summary, factorization into real number factors is possible when \( \Delta \geq 0 \), with a single repeated factor for \( \Delta = 0 \), and two distinct factors for \( \Delta > 0 \). For \( \Delta < 0 \), factorization into real factors is not possible.

Examples

The following example should make it clear how to apply the above steps.

Let us factorize \( -2x^2 + 4x + 3 \). Observe that \(\Delta=b^2-4ac=(4)^2-4(-2)(3)\)\(>0\). This means that the given expression is factorizable into linear factors.

  1. Write the expression: \( -2x^2 + 4x + 3 \).

  2. Factor out \(-2\), keeping it with the expression: \( -2(x^2 - 2x - \frac{3}{2}) \).

  3. Complete the square inside the parenthesis: \( x^2 - 2x \) becomes \( (x - 1)^2 \) by adding and subtracting \( 1 \). So, \( -2(x^2 - 2x - \frac{3}{2}) \) becomes \( -2((x - 1)^2 - 1 - \frac{3}{2}) \).

  4. Simplify inside the parenthesis: \( (x - 1)^2 - 1 - \frac{3}{2} = (x - 1)^2 - \frac{5}{2} \).

  5. The factorized form is: \( -2((x - 1)^2 - \frac{5}{2}) \).

May be another example: Factorize \( 3x^2 - 5x + 1 \). For this example as well \(\Delta>0\)

To factorize \( 3x^2 - 5x + 1 \) by completing the square, follow these steps:

  1. Write the expression: \( 3x^2 - 5x + 1 \).

  2. Factor out the coefficient of \( x^2 \), which is \( 3 \): \( 3(x^2 - \frac{5}{3}x + \frac{1}{3}) \).

  3. Complete the square inside the parenthesis: \( x^2 - \frac{5}{3}x \) becomes \( \left(x - \frac{5}{6}\right)^2 \) by adding and subtracting \( \left(\frac{5}{6}\right)^2 \). So, \( 3(x^2 - \frac{5}{3}x + \frac{1}{3}) \) becomes \( 3\left(\left(x - \frac{5}{6}\right)^2 - \left(\frac{5}{6}\right)^2 + \frac{1}{3}\right) \).

  4. Simplify inside the parenthesis: \( \left(x - \frac{5}{6}\right)^2 - \left(\frac{5}{6}\right)^2 + \frac{1}{3} = \left(x - \frac{5}{6}\right)^2 - \frac{25}{36} + \frac{12}{36} = \left(x - \frac{5}{6}\right)^2 - \frac{13}{36} \).

  5. The factorized form is: \( 3\left(\left(x - \frac{5}{6}\right)^2 - \frac{13}{36}\right) \).

Let us look at an example in which \(\Delta=0\). Factorize \(4x^2+12x+9\)

  1. Take 4 common: \( 4x^2 + 12x + 9 = 4(x^2 + 3x + \frac{9}{4}) \)

  2. Complete the Square for the expression inside the parentheses: \( x^2 + 3x + \frac{9}{4} \)
    To complete the square, add and subtract the square of half the coefficient of \(x\). The coefficient of \(x\) is 3, so half of it is \(\frac{3}{2}\), and its square is \(\left(\frac{3}{2}\right)^2 = \frac{9}{4}\).
    \( x^2 + 3x + \frac{9}{4} = \left(x + \frac{3}{2}\right)^2 \)

  3. Substitute back into the original expression: \( 4(x^2 + 3x + \frac{9}{4}) = 4\left(x + \frac{3}{2}\right)^2 \)\(=\left(2x + 3\right)^2\)

Therefore, the factorized form of \(4x^2 + 12x + 9\) is \(\left(2x + 3\right)^2\).

Now, let us look at an example in which \(\Delta<0\). Factorize \(3x^2 - 5x + 5\).

  1. Take 3 common: \( 3x^2 - 5x + 5 = 3(x^2 - \frac{5}{3}x + \frac{5}{3}) \)

  2. Complete the Square for the expression inside the parentheses: \( x^2 - \frac{5}{3}x \)
    To complete the square, add and subtract the square of half the coefficient of \(x\). The coefficient of \(x\) is \(-\frac{5}{3}\), so half of it is \(-\frac{5}{6}\), and its square is \(\left(-\frac{5}{6}\right)^2 = \frac{25}{36}\).
    \( = x^2 - \frac{5}{3}x + \frac{25}{36} - \frac{25}{36} + \frac{5}{3} \)
    \( = x^2 - \frac{5}{3}x + \frac{25}{36} = \left(x - \frac{5}{6}\right)^2 \)
    \( = 3\left(\left(x - \frac{5}{6}\right)^2 - \frac{25}{36} + \frac{5}{3}\right) \)

  3. Simplify: \( 3\left(\left(x - \frac{5}{6}\right)^2 + \frac{45}{36} - \frac{25}{36}\right) \) \( =3\left(\left(x - \frac{5}{6}\right)^2 + \frac{20}{36}\right) \)

  4. Examine for Factorization: Since the term inside the parentheses is a sum of squares \(\left(\left(x - \frac{5}{6}\right)^2 + \frac{20}{36}\right)\), it cannot be further factorized using real numbers.

Therefore, the expression \(3x^2 - 5x + 5\) is irreducible over the real numbers.

Irreducibility

A quadratic expression is said to be irreducible over the real numbers \( \mathbb{R} \) if it cannot be factored into the product of two linear expressions with real coefficients. In simpler terms, an irreducible quadratic expression over \( \mathbb{R} \) does not have any real roots.

Mathematically, a quadratic expression of the form:

\[ ax^2 + bx + c \]

is irreducible over \( \mathbb{R} \) if there are no real numbers \( p \), \( q \), \( r \), and \( s \) such that:

\[ ax^2 + bx + c = (px + q)(rx + s) \]

for all \( x \) in \( \mathbb{R} \).

Determining Irreducibility

The irreducibility of a quadratic expression over \( \mathbb{R} \) is typically determined by examining its discriminant, \( \Delta \), where \( \Delta = b^2 - 4ac \). The expression is irreducible over \( \mathbb{R} \) if and only if \( \Delta < 0 \). When the discriminant is negative, the quadratic equation \( ax^2 + bx + c = 0 \) does not have real solutions, and thus the expression cannot be factored into real linear factors.

Example

For instance, the quadratic expression \( x^2 + 1 \) is irreducible over \( \mathbb{R} \) since its discriminant \( \Delta = 0^2 - 4 \cdot 1 \cdot 1 = -4 \) is less than zero. Therefore, there are no real numbers \( x \) that satisfy \( x^2 + 1 = 0 \), and the expression cannot be factored into linear terms with real coefficients.

Roots(🌳)

Now that we understand how to factorize a quadratic expression, we look at what roots of a quadratix expressions are. Simply, the roots of a quadratic expression are the solutions to the quadratic equation \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are real numbers and \( a \neq 0 \). The roots are the values of \( x \) for which the value of the expression becomes exactly zero. That's why these values of \(x\) are also called zeros. The rocess again involves completing the square method.

So, we start with quadratic eqaution \( ax^2 + bx + c = 0 \) and follow these steps. ➡Step 1. Rearranging the Equation: Divide the entire equation by \( a \) (if \( a \) is not 1) to make the coefficient of \( x^2 \) equal to 1: [ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 ]

➡Step 2. Moving the Constant Term: Shift the constant term to the right side: [ x^2 + \frac{b}{a}x = -\frac{c}{a} ]

➡Step 3. Completing the Square: Add \(\left(\frac{b}{2a}\right)^2\) (the square of half the coefficient of \( x \)) to both sides: [ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a} ] This forms a perfect square on the left side: [ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} ]

➡Step 4: Introducing Discriminant (\( \Delta \)):

Define \( \Delta = b^2 - 4ac \). The equation becomes: [ \left(x + \frac{b}{2a}\right)^2 = \frac{\Delta}{4a^2} ]

🔍 Analysis Based on Discriminant (\( \Delta \))

🔘 Case I: For \( \Delta > 0 \):

  • The equation has two real and distinct roots.
  • Solve using: \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \).

🔘 Case II: For \( \Delta = 0 \):

  • The equation has one real root (a repeated root).
  • The root is \( x = \frac{-b}{2a} \).

🔘 Case III. For \( \Delta < 0 \):

  • The equation has no real roots. This is because the left-hand side of the equation, \(\left(x + \frac{b}{2a}\right)^2\), represents a square and thus is always non-negative (as the square of any real number, whether positive, negative, or zero, is non-negative).
  • However, for \( \Delta < 0 \), the right-hand side, \(\frac{\Delta}{4a^2}\), is negative (since \( \Delta \) is negative and \( a^2 \) is always non-negative).
  • Since a non-negative number cannot equal a negative number, there are no real values of \( x \) that satisfy the equation when \( \Delta < 0 \).

Roots and Factorization

The factorization of a quadratic expression \( ax^2 + bx + c \) is intimately connected to its roots. When this expression is factorized, it takes the form \( a(x - \alpha)(x - \beta) \), where \( \alpha \) and \( \beta \) represent the roots of the quadratic equation \( ax^2 + bx + c = 0 \). These roots are the specific values of \( x \) that satisfy the equation, making the entire expression equal to zero.

The roots, \( \alpha \) and \( \beta \), are determined using the quadratic formula: \( \alpha, \beta = \frac{-b \pm \sqrt{\Delta}}{2a} \), where \( \Delta \) is the discriminant given by \( b^2 - 4ac \). This formula provides the solutions to the quadratic equation, and these solutions are precisely the values at which the factorized expression \( a(x - \alpha)(x - \beta) \) equals zero.

In essence, the process of factorizing the quadratic expression reveals the roots directly. Each factor in the expression \( a(x - \alpha)(x - \beta) \) corresponds to a root of the quadratic equation. When \( x \) takes the value of \( \alpha \) or \( \beta \), the corresponding factor becomes zero, thereby making the product of the factors, and hence the entire expression, equal to zero. This relationship between factorization and the roots of a quadratic equation is fundamental in algebra, allowing for an understanding of how the solutions to the equation are intrinsically linked to the expression's factorized form.

Vieta's Formula:

Vieta's formulas for a quadratic expression provide a powerful relationship between the roots of the expression and its coefficients. These formulas state that for a quadratic expression of the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a \neq 0 \), the sum and product of its roots are directly related to the coefficients of the expression.

The Quadratic Expression and Its Roots

Consider the quadratic expression \( ax^2 + bx + c \). The roots of this expression, let's call them \( \alpha \) and \( \beta \), are the solutions to the equation \( ax^2 + bx + c = 0 \). These roots can be real or complex numbers depending on the discriminant of the quadratic equation.

Vieta's Formulas

The fundamental idea behind Vieta's formulas comes from the fact that a quadratic expression with roots \( \alpha \) and \( \beta \) can be factorized as \( a(x - \alpha)(x - \beta) \). By expanding this factorized form, we get:

\[ a(x - \alpha)(x - \beta) = ax^2 - a(\alpha + \beta)x + a\alpha\beta \]

Now, compare this expanded form to the original expression \( ax^2 + bx + c \). By equating the coefficients of the corresponding terms of \( x^2 \), \( x \), and the constant term, we derive the formulas:

  1. Sum of the Roots: Equating the coefficients of \( x \) from \( -a(\alpha + \beta)x \) and \( bx \) gives us the first of Vieta's formulas: \( \alpha + \beta = -\frac{b}{a} \).
    This formula states that the sum of the roots of the quadratic expression is equal to the negative of the coefficient of \( x \) divided by the coefficient of \( x^2 \).

  2. Product of the Roots: Equating the constant terms \( a\alpha\beta \) and \( c \) gives us the second formula: \( \alpha\beta = \frac{c}{a} \).
    This means that the product of the roots of the quadratic expression is equal to the constant term divided by the coefficient of \( x^2 \).

Thus, Vieta's formulas establish a clear and direct relationship between the coefficients of a quadratic expression and the roots of the corresponding quadratic equation. The elegance of these formulas lies in their ability to provide information about the roots without necessarily having to solve the equation explicitly. They are particularly useful in various mathematical applications, including algebraic manipulation and problem-solving.

When discriminant is negative.

One thing which is not clear now is what happens when discriminant is negative. As negative discriminant indicates that quadratic expression does not have real roots, does the sum and product of roots still make sense? Since, we have not yet defined the concept of complex numbers, Vieta's formulas clearly are absurd in this case. But once we define complex numbers, these formulas will absolutely make sense for negative discriminant.

More than two roots 🤔

If a quadratic expression has three distinct roots (let's call them \( \alpha \), \( \beta \), and \( \gamma \)), then the coefficients of the quadratic expression must be zero (i.e., \( a = 0 \), \( b = 0 \), and \( c = 0 \)). To prove this, we start with the standard form of a quadratic expression:

\[ ax^2 + bx + c \]

Given that \( \alpha \), \( \beta \), and \( \gamma \) are roots of this expression, it means that the expression equals zero when \( x \) is replaced by each of these roots. Therefore, we have:

  1. \( a\alpha^2 + b\alpha + c = 0 \) [1]
  2. \( a\beta^2 + b\beta + c = 0 \) [2]
  3. \( a\gamma^2 + b\gamma + c = 0 \) [3]

⚙ First, subtract [2] from [1]:

\( a\alpha^2 + b\alpha + c - (a\beta^2 + b\beta + c) = 0 - 0 \)
\(\implies\)\( a(\alpha^2 - \beta^2) + b(\alpha - \beta) = 0 \)
\(\implies\)\( (\alpha - \beta)(a(\alpha + \beta) + b) = 0 \)

Since \( \alpha \neq \beta \) (the roots are distinct), \( \alpha - \beta \neq 0 \). Therefore, the other factor must be zero:
\( a(\alpha + \beta) + b = 0 \) [4]

⚙ Similarly, subtract [3] from [1]:

\( a(\alpha^2 - \gamma^2) + b(\alpha - \gamma) = 0 \)
\(\implies\)\( (\alpha - \gamma)(a(\alpha + \gamma) + b) = 0 \)

Since \( \alpha \neq \gamma \), \( \alpha - \gamma \neq 0 \). Thus:
\( a(\alpha + \gamma) + b = 0 \) [5]

⚙ Subtract Equation [4] from [5]

\( a(\alpha + \gamma) + b - (a(\alpha + \beta) + b) = 0 - 0 \)
\(\implies\)\( a\gamma - a\beta = 0 \)
\(\implies\)\( a(\gamma - \beta) = 0 \)

Since we assume \( a \neq 0 \), it follows that \( \gamma - \beta = 0 \) or \( \beta = \gamma \), which contradicts our initial assumption of three distinct roots.

This contradiction implies that our initial assumption of having three distinct roots for a quadratic expression is flawed. Therefore, the only way a quadratic expression can have three distinct roots is if \( a = 0 \), \( b = 0 \), and \( c = 0 \), which essentially means the expression is identically zero. This aligns with the fundamental nature of a quadratic equation, which can have at most two distinct roots.

Newton's identities

For a quadratic equation \( ax^2 + bx + c = 0 \) with roots \( \alpha \) and \( \beta \), Newton's identity states that for any natural number \( n \) greater than or equal to 2, the following recursive relationship holds:

\[ aS_n + bS_{n-1} + cS_{n-2} = 0 \]

where \( S_n = \alpha^n + \beta^n \).

Proof:

  1. Begin with the Roots: Since \( \alpha \) and \( \beta \) are roots of the quadratic equation, they satisfy:

    \( a\alpha^2 + b\alpha + c = 0 \) [1]
    \( a\beta^2 + b\beta + c = 0 \) [2]

  2. Multiply by Powers of \( \alpha \) and \( \beta \): Multiply equation [1] by \( \alpha^{n-2} \) and equation [2] by \( \beta^{n-2} \):

    \( a\alpha^n + b\alpha^{n-1} + c\alpha^{n-2} = 0 \) [3]
    \( a\beta^n + b\beta^{n-1} + c\beta^{n-2} = 0 \) [4]

  3. Add the Equations: Add equations [3] and [4] to get:

    \( a(\alpha^n + \beta^n) + b(\alpha^{n-1} + \beta^{n-1}) + c(\alpha^{n-2} + \beta^{n-2}) = 0 \)

    Simplifying this, we have: \( aS_n + bS_{n-1} + cS_{n-2} = 0 \)

  4. Initial Cases: For \( n = 1 \) and \( n = 2 \), the initial cases are:

    \( S_1 = \alpha + \beta = -\frac{b}{a} \)
    \( S_2 = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{b}{a}\right)^2 - 2\frac{c}{a} \)

This recursive nature of Newton's identities provides a method for calculating higher power sums of the roots without directly solving for the roots themselves.

Difference of Roots

Given a quadratic equation \( ax^2 + bx + c = 0 \) with roots \( \alpha \) and \( \beta \), the absolute difference between the roots is given by the formula:

\[ |\alpha - \beta| = \frac{\sqrt{D}}{|a|} \]

where \( D = b^2 - 4ac \) is the discriminant of the quadratic equation.

Proof:

By Vieta's formulas, the sum and product of the roots \( \alpha \) and \( \beta \) of the quadratic equation are given as follows:

  • The sum of the roots \( \alpha + \beta = -\frac{b}{a} \).
  • The product of the roots \( \alpha\beta = \frac{c}{a} \).

We aim to find the absolute value of the difference of the roots, \( |\alpha - \beta| \). To do this, we start with the identity that relates the square of the difference of the roots to the sum and product of the roots:

\( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \)

Substituting Vieta's formulas into the identity gives:

\( (\alpha - \beta)^2 = \left(-\frac{b}{a}\right)^2 - 4\left(\frac{c}{a}\right) \)
\(\implies\)\( (\alpha - \beta)^2 = \frac{b^2}{a^2} - \frac{4ac}{a^2} \)
\(\implies\)\( (\alpha - \beta)^2 = \frac{b^2 - 4ac}{a^2} \)

Recognizing that \( b^2 - 4ac \) is the discriminant \( D \), we can write:

\[ (\alpha - \beta)^2 = \frac{D}{a^2} \]

Taking the square root of both sides, and considering the absolute value to account for the order of subtraction in the roots, we obtain:

\[ |\alpha - \beta| = \frac{\sqrt{D}}{|a|} \]

This completes the proof. The absolute difference between the roots of the quadratic equation \( ax^2 + bx + c = 0 \) is indeed \( \frac{\sqrt{D}}{|a|} \), where \( D \) is the discriminant of the equation.