# 38. Quadratic Equations¶

An equation of the form \(ax^2 + bc + c = 0\) where \(a, b, c\) are real numbers, and \(a \ne 0\) is called
a *quadratic equation*.

The numbers \(a, b, c\) are called the coefficients of the quadratic equation and \(b^2 - 4ac\) is called its discriminant.

Discriminant of a quadratic equation is donated by \(D\) or \(\Delta\).

**Examples:**

- \(4x^2 + 4x + 1 = 0, a = 4, b = 4, c = 1\)
- \(7x^3 + 10 = 0\) is not a quadratic equation.
- \(3x^2 -2x^{1/2} + 7 = 0\) is not a quadratic equation.
- \(2x^2 - 4 = 0, a = 2, b = 0, c= -4\)

The quadratic equation is called incomplete if one of the coefficients \(b\) or \(c\) is zero. Thus, the last example above represents an incomplete quadratic equation.

An expression of the form \(ax^2 + bx + c\) is called a *quadratic expression* while other elements are same as a
quadratic equation.

If two expressions in \(x\) are equal for all values of \(x\) then this statement of equality between the two expressions is called an identity.

\(f(x) = 0\) is said to be an identity in \(x\) if it is satisfied by all values of \(x\) in the domain of \(f(x)\).

Thus, an identity is satisfied by all values of \(x\) while an equation is satisfied for particular values of \(x\).

**Example:**

- \((x + 1)^2 = x^2 + 2x + 1\) is an identity in \(x\)

Two equations are called *identical equations* if they have same roots.

**Example:** \(x^2 - 5x + 4 = 0\) and \(2x^2 - 10x + 8 = 0\) are identical equations because both have same
roots \(1\) and \(4\).

**Note:**

- Two equations in \(x\) are identical if and only if the coefficients of similar power of \(x\) in the two equations are proportional. Thus, if \(ax^2 + bx + c = 0\) and \(a_1x^2 + b_1x + c_1 = 0\) are identical equations, then \(\frac{a}{a_1} = \frac{b}{b_1} = \frac{c}{c_1}\)
- An equation remains unchanged if it is multiplied or divided by a non-zero number.

An expression of the form \(a_0x^n + a_1x^{n - 1} + a_2x^{n - 2} + ... + a_{n - 1}x + a_0\) where \(a_0, a_1, a_2, ..., a_n\) are constants (\(a_0 \ne 0\)) and \(n\) is a positive integer is called a polynomial in \(x\) of degree \(n\).

As a special case a constant is also called a polynomial of degree zero.

**Example:** \(8x^7 -6x^5 + 3x^2 + x + 9\) is a polynomial of degree \(7\) but \(x + \frac{1}{x}\) is not a
polynomial in \(x\).

## 38.1. Rational expression or Rational function¶

An expression of the form \(\frac{P(x)}{Q(x)},\) where \(P(x)\) and \(Q(x)\) are polynomials in \(x\), is called a rational expression.

In the particular case when \(Q(x)\) is a non-zero constant, \(\frac{P(x)}{Q(x)}\) reduces to a polynomial. Thus every polynomial is a rational expression but the converse is not true.

**Examples:**

- \(\frac{x^2 - 5x + 4}{x - 2}\)
- \(\frac{1}{x - 7}\)

## 38.2. Roots of a quadratic equation¶

The values of \(x\) for which the equation \(ax^2 + bx + c = 0\) are satisfied are called roots of the equation. They are also called roots of the quadratic expression \(ax^2 + bx + c\).

Every quadratic equation has at most two roots. Proof is given below:

Let \(ax^2 + bx + c = 0\) where \(a \ne 0\)

Multiplying both sides of the equation by \(a\)

\(a^2x^2 + 2abx + ac = 0 \Rightarrow (ax)^2 + 2.ax.\frac{b}{2} + \left(\frac{b}{2}\right)^2 + ac - \left(\frac{b}{2}\right)^2 = 0\)

\(\left(ax + \frac{b}{2}\right)^2 = \frac{b^2 - 4ac}{4}\)

\(ax + \frac{b}{2} = \pm\frac{\sqrt{b^2 - 4ac}}{2}\)

\(\therefore a = \frac{-b \pm \sqrt{b^2 - 2ac}}{2a}\)

These are two roots of a quadratic equation.

Let us suppose the above quadratic equation has three roots \(\alpha, \beta\) and \(\gamma\). These roots will satisfy the equation

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

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

\(a\gamma^2 + b\gamma + c = 0\)

Subtracting first two we get

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

\(\because \alpha \ne \beta \therefore a(\alpha + \beta) + b = 0\)

Similarly, \(a(\alpha + \gamma) + b = 0\)

Subtracting these two we get \(a(\alpha - \gamma) = 0\)

Since \(a\ne 0 \therefore \alpha = \gamma\)

Thus a quadratic equation has at most two roots.

## 38.3. Sum and product of the roots¶

If \(\alpha\) and \(\beta\) are two roots of the equation \(ax^2 + bx + c = 0\) then

\(\alpha = \frac{-b + \sqrt{b^2 - 4ac}}{2a}\) and \(\beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\)

\(\alpha + \beta = -\frac{b}{a} = - \frac{\text{coeff. of } x}{\text{coeff. of }x^2}\)

\(\alpha\beta = \frac{c}{a} = \frac{\text{constant term}}{\text{coeff. of }x^2}\)

## 38.4. Discriminant of the quadratic equation¶

\(b^2 - 4ac\) is called the discriminant of the quadratic equation and is usually denoted by \(D\) or \(\Delta\).

## 38.5. Nature of roots¶

For equation \(ax^2 + bx + c = 0\) when \(a, b, c\) are real.

When \(D < 0\)

In this case both roots \(\alpha\) and \(\beta\) will be either complex numbers or imaginary depending on whether \(b\) is zero or not as discriminant is imaginary. These roots are conjugate of each other which you can verify easily.

When \(D = 0\)

In this case both roots will be equal.

When \(D > 0\)

In this case both the roots will be equal and unequal. If \(D\) is not a perfect square(square of a rational number) then roots are irrational and come as a pair of conjugate irrational numbers which you can verify easily.

When \(D\) i.e. \(b^2 - 4ac\) is a perfect square(square of a rational number) and \(a, b, c\) are rationals.

In this case \(b^2 - 4ac\) = square of a rational number

\(\therefore \sqrt{b^2 - 4ac}\) = a rational number, let \(\sqrt{b^2 - 4ac} = k\)

Thus, \(\alpha = \frac{-b - k}{2a}\) and \(\beta = \frac{-b + k}{2a}\) where \(a, b, k\) are rationals.

### 38.5.1. Conjugate Roots¶

Imaginary roots of a quadratic equation with real coefficients always occur in conjugate pair.

Let \(\alpha + i\beta\) be a root of the quadratic equation \(ax^2 + bx + c = 0\) where \(a, b, c\) are real numbers. Thus,

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

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

Equating real and imaginary parts

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

Using \(\alpha - i\beta\) as the second root of the equation

\(a(\alpha - i\beta)^2 + b(\alpha - i\beta) + c\)

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

\(= 0 + i.0\)

Thus, we can see that \(\alpha -i\beta\) also satisfies the equation and is second root of the equation.

### 38.5.2. Irrational Roots¶

Like imaginary roots, irrational roots also appear in pair as conjugate roots of a quadratic equation. Proof has been left as an exercise to the reader.

## 38.6. Symmetric functions of roots¶

If a function of \(\alpha\) and \(\beta\) remain unchanged when they are interchanged then the function is called symmetric function of \(\alpha\) and \(\beta\). For example, \(\alpha^2 + \beta^2 + \alpha\beta\) is a symmetric function while \(\alpha^2 + \beta^2 + \alpha\) is not a symmetric function.

## 38.7. Representing the equation in terms of roots¶

Let \(ax^2 + bx + c = 0\) be a quadratic equation whose roots are \(\alpha\) and \(\beta\).

\(ax^2 + bx + c = x^2 + \left(\frac{b}{a}\right) + \left(\frac{c}{a}\right)\)

\(x^2 - \left(-\frac{b}{a}\right) + \frac{c}{a} = x^2 -(\alpha + \beta)x + \alpha\beta = 0\)

## 38.8. Condition for common roots¶

Let \(ax^2 + bx + c = 0\) and \(a_1x^2 + b_1x + c_1 = 0\) have a common root.

Let \(c, c_1 \ne 0\) and \(ab_1 - a_1b \ne = 0\). Let the common root be \(\alpha\) then.

\(a\alpha^2 + b\alpha + c = 0\) and \(a_1\alpha^2 + b_1\alpha + c_1 = 0\)

By cross-multiplication

\(\frac{\alpha^2}{bc_1 - b_1c} = \frac{\alpha}{ca_1 - c_1a} = \frac{1}{ab_1 - a_1b}\)

\(\Rightarrow (bc_1 - b_1c)(ab_1 - a_1b) = (ca_1 - c_1a)^2\)

This is the required condition.

**Note.** If \(c = c_1 = 0,\) then equations \(ax^2 + bx + c = 0\) and \(a_1x^2 + b_1x + c_1 = 0\) will
reduce to \(ax^2 + bx = 0\) and \(a_1x^2 + bx_1 = 0\) and have \(0\) as a common root. The other roots would
be \(-\frac{b}{a}\) and \(-\frac{b_1}{a_1}\). Thus, if \(a_1b = ab_1\) then both the roots would be common.

For having both the roots common the equations must be identical i.e. \(\frac{a}{a_1} = \frac{b}{b_1} = \frac{c}{c_1}\)

## 38.9. Sign of quadratic expression \(ax^2 + bx + c\)¶

Let \(y = ax^2 + bx + c\) and let \(\alpha\) and \(\beta\) be the root of the quadratic expression. Then, \(\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a}\) and \(ax^2 + bx + c = a(x - \alpha)(x - \beta)\)

\(\therefore y = ax^2 + bx + c\)

Case I: When \(\alpha\) and \(\beta\) are complex numbers.

Let \(\alpha = p + iq\), then \(\beta = p - iq\) where \(q \ne 0\)

\(ax^2 + bx + c = a[x - (p + iq)][x - (p - iq)]\)

\(= a[(x - p)^2 + q^2] = a~\times\) a positive quantity

\(ax^2 + bx + c\) will have same sign as that of \(a\) for all real \(x\).

Case II: When \(\alpha\) and \(\beta\) are real and equal.

Given \(\alpha = \beta\)

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

Thus, the expression will have same sign as \(a\) except when \(x = \alpha\) in which case it will be 0.

Case III: When \(\alpha\) and \(\beta\) are real and unequal.

Sub case (i) When \(x < \alpha < \beta\)

\(\because x < \alpha \Rightarrow x - \alpha < 0\) and \(x < \beta \Rightarrow x - \beta < 0\)

\(\therefore a(x - \alpha)(x - \beta) > 0\) thus the expression \(ax^2 + bx + c\) will have same sign as that of \(a\).

Sub case (ii) When \(x > \alpha > \beta\)

\(\because x > \alpha \Rightarrow x - \alpha > 0\) and \(x > \beta \Rightarrow x - \beta > 0\)

\(\therefore a(x - \alpha)(x - \beta) > 0\) thus the expression \(ax^2 + bx + c\) will have same sign as that of \(a\).

Sub case (iii) When \(\alpha < x < \beta\)

\(\because x > \alpha \Rightarrow x - \alpha > 0\) and \(x < \beta \Rightarrow x - \beta < 0\)

\(\therefore a(x - \alpha)(x - \beta) < 0\) thus the expression \(ax^2 + bx + c\) will have opposite sign as that of \(a\).

## 38.10. Maximum and minimum values of \(ax^2 + bx + c\)¶

Let \(y = ax^2 + bx + c \Rightarrow ax^2 + bx + c - y = 0\)

Since \(x\) is real, therefore, discriminant has to be greater than 0.

\(\therefore b^2 - 4a(c - y) \geq 0 \Rightarrow b^2 - 4ac + 4ay \geq 0\)

\(\Rightarrow y \geq \frac{4ac - b^2}{4a}\)

**Case I:** When \(a > 0\)

Clearly, minimum value of \(y\) is \(\frac{4ac - b^2}{4a}\).

Substituting this for \(ax^2 + bx + c = y\) and solving we see that it occurs for \(x = -\frac{b}{2a}\)

Thus, minimum value of \(y = \frac{4ac - b^2}{4a}\) and it has no maximum value.

**Case II:** When \(a < 0\)

Clearly, when \(a < 0, y\) has no minimum value and maximum value will again occur at \(x = -\frac{b}{2a}\).

## 38.11. To find the condition that the general quadratic equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c\) in \(x\) and \(y\) may be resolved into two linear rational factors.¶

Corresponding equation is \(ax^2 + 2hxy + by^2 + rgx + 2fy + c = 0\)

\(x = \frac{-2(hy + g)\pm\sqrt{4(hy + g)^2 - 4a(by^2 + 2fy + c)}}{2a}\)

\(ax + hy + g = \pm\sqrt{(h^2 - ab)y^2 + 2(gh - af)y + g^2 - ac}\)

It can be resolved into two linear factors if \((h^2 - ab)y^2 + 2(gh - af)y + g^2 - ac\) is a perfect square and \(h^2 - ab > 0\)

\((h^2 - ab)y^2 + 2(gh - af)y + g^2 - ac\) will be prefect square if discriminant of the corresponding equation is \(0\).

\(\Rightarrow 4(gh - af)^2 - 4(h^2 - ab)(g^2 - ac) = 0\)

\(\Rightarrow abc + 2fgh - af^2 - bg^2 - ch^2 = 0\)