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:

1. $$4x^2 + 4x + 1 = 0, a = 4, b = 4, c = 1$$

2. $$7x^3 + 10 = 0$$ is not a quadratic equation.

3. $$3x^2 -2x^{1/2} + 7 = 0$$ is not a quadratic equation.

4. $$2x^2 - 4 = 0, a = 2, b = 0, c= -4$$

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:

1. $$(x + 1)^2 = x^2 + 2x + 1$$ 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:

1. 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 $$\frac{a}{a_1} = \frac{b}{b_1} = \frac{c}{c_1}$$

2. 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 $$x$$.

## 45.1. 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:

1. $$\frac{x^2 - 5x + 4}{x - 2}$$

2. $$\frac{1}{x - 7}$$

## 45.2. 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.

## 45.3. 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 $$\beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$$

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

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

## 45.4. Discriminant of the quadratic equation

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

## 45.5. Nature of roots

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

1. 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.

2. When $$D = 0$$

In this case both roots will be equal.

3. 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.

4. 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.

### 45.5.1. 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.

### 45.5.2. 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.

## 45.6. 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.

## 45.7. Representing the equation in terms of roots

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

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

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

## 45.8. 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. $$\frac{a}{a_1} = \frac{b}{b_1} = \frac{c}{c_1}$$

## 45.9. 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 $$a$$.

## 45.10. 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 $$x = -\frac{b}{2a}$$.

## 45.11. 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 $$0$$.

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

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