**7.1. Introduction**

The equation is one of the most prominent ideas in mathematics, and is the center of foundation of mathematics itself. Beginning in the “Before Christ” era, the Babylonians were the first to have been recorded demonstrating the equation, circa 400 BC.Throughout the years, the history of mathematics has taken its fair share of changes. It has stretched across the world from the Far East, migrating into the Western Hemisphere. One of the most fundamental and key principles of mathematics has been the quadratic formula. Having been used in several different cultures, the formula has been part of the base of mathematics theory.

Pacioli never actually devised his own method of solving the quadratic equation, but instead published a works dealing with the history of certain aspects of mathematics, including idea of all the great mathematicians.

Overall, the quadratic formula has actually shaped civilization into the way it is today. The Babylonians used it to irrigate their land, divide out funds to pay workers, and even mobilize armies. The formula to them was not a way of learning or teaching, but a way of life. Before we know further, let us know some pre-requisites of the chapter.

**Polynomial**

It is an expression in which powers of variables are non-negative integers.

The general form of a polynomial is

${a}_{n}$${x}^{n}$ + ${a}_{n\u20131}$${x}^{n\u20131}$ + ${a}_{n\u20132}$${x}^{n\u20132}$ + – – – – +${a}_{1}$x+ ${a}_{0}$

where ${a}_{n}$, ${a}_{n\u20131}$, ${a}_{n\u20132}$ – – – – ${a}_{0}$ are real numbers and n is a positive integer.

**Examples:** 2009${x}^{3}$ + 2008 ${x}^{2}$ + 2007x + 2006

$\frac{1}{2}{x}^{10}+\sqrt{2}{x}^{5}+0.25x+2009$

Degree of polynomial **Polynomial of one variable :** In the case of a polynomial in one variable, the highest power of the variable is called the degree of the polynomial.**Example:**

i) 2009${z}^{2009}$+ 2008${z}^{2008}$ + 2007${z}^{2007}$ + ….. + ${z}^{2}$ + z + 1 is a polynomial in z of degree 2009.

ii) 10${x}^{5}$ + 2${x}^{4}$ + 3${x}^{3}$ + 4${x}^{2}$ + 5x + 6

The term with the highest power of x is 10${x}^{5}$.

The degree of the given polynomial is ‘5’. **Polynomial of several variables :** In the case of polynomials in more than one variable, the sum of the powers of the variables in each term is taken up and the highest sum so obtained is called the degree of the polynomial.**Example:**

257${a}^{2}$${b}^{5}$${c}^{7}$ – 342 ${a}^{3}$ ${b}^{4}$ ${c}^{2}$ + 532 ${a}^{5}$${b}^{3}$${c}^{2}$ – 198 a${b}^{9}$${c}^{8}$

the highest sum of the variables is 18.

$\therefore $ The degree of a polynomial is 18 in a, b and c.

2${x}^{2}$yz + 5${x}^{2}$${y}^{2}$${z}^{2}$ – 7${x}^{3}$${y}^{2}$${z}^{2}$ – ${x}^{3}$${y}^{3}$${z}^{3}$

The sum of the powers of the variables in the terms of the above polynomial are 4, 6, 7 and 9 respectively. The highest power is 9. Thus, the degree of the given polynomial is 9.

**Quadratic Polynomial**

A polynomial with degree two is called a quadratic polynomial.The general form of a quadratic polynomial is

a${x}^{2}$ + bx + c, where $a\ne 0$ (compulsory condition) and a,b,c$\in $R**Example :** ${x}^{2}$ + 2, 2${x}^{2}$ + 3x – 4,${x}^{2}\u2013\sqrt{2}x$,$\sqrt{3}{x}^{2}+x\u20137$ etc.

We denote quadratic polynomial by p(x), q(x), f(x) etc.**Zeros of a quadratic polynomial**

Let us now consider a quadratic polynomial in x, say P(x) = ${x}^{2}$ + x – 2; we randomly substitute some real values of x and find the corresponding values of P(x)

x P(x)

–2 0

–1 –2

0 –2

1 0

2 4

We observe that P(x) is zero only when x is –2 and 1 and these values are called zeros, roots or solutions of the polynomial. **Definition:** The values of a variable, for which a polynomial p(x) becomes zero, are called zeros of a polynomial.

If a is zero of a polynomial p(x), then p(a) = 0.

**Points to remember**

i. It is a fact that every quadratic polynomial can have at most two zeros.**For example**, ${x}^{2}$ + 5x + 6 = 0 there are two zeros and they are –2 , –3

ii. Some quadratic polynomials do not have any zero. That is to say, there is no real value of x which makes the value of the polynomial zero. **For example**, consider the quadratic polynomial ${x}^{2}$ + 1. There is no real value of x which makes ${x}^{2}$ + 1 zero, since for every real value of x, x20 which implies that ${x}^{2}$ + 1 > 0.

iii. Similarly, some quadratic polynomials have only one zero, i.e., there is only one real value of x which makes the value of the polynomial zero. **For example**, consider the quadratic polynomial. ${x}^{2}$ – 2x + 1.There is only one value of x(x = 1), for which ${x}^{2}$ – 2x + 1 is zero.

**7.2. Quadratic Equaation**

A second degree equation is called a Quadratic Equation.

or

An equation whose highest power of a variable is two is known as a Quadratic Equation.

The general form of a quadratic equation is

a${x}^{2}$ + bx + c = 0, where $a,b,c\in R$ a$\ne $0 (compulsory condition).

Since x$\ne $0, quadratic equations, in general, are of the following types:

i) b = 0, c$\ne $0, i.e., of the type a${x}^{2}$ + c = 0**Example:** 5${x}^{2}$ = 125

ii) b$\ne $0, c = 0, i.e., of the type a${x}^{2}$ + bx = 0**Example:** 2${x}^{2}$ + 3x = 0

iii) b = 0, c = 0, i.e., of the type a${x}^{2}$ = 0**Example:** 6${x}^{2}$ = 0

iv) b$\ne $0, c$\ne $0, i.e., of the type a${x}^{2}$ + bx + c = 0**Example: **${x}^{2}$ + 5x + 4 = 0

**Pure Quadratic Equations**

The quadratic equation $a{x}^{2}+bx+c=0(a\ne 0)$ is called a pure quadratic equation if the x coefficient is 0 i.e., b = 0.**Example : $3{x}^{2}+9=0$** **Monic Quadratic Equations**

The quadratic equation ${\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}=0(\mathrm{a}\ne 0)$ is called a monic quadratic equation if the coefficient of its leading term is 1 i.e., a = 1.**Example: ${x}^{2}\u20137x+12=0$** **Note:** Highest degree term in an equation is called its leading term.**Root of an Equation**

The value of the variable satisfying the given equation is called its root.

In the sense, x = $\propto $ is a root of the equation f(x) = 0 if and only if $f\left(\propto \right)=0$

Example.

Consider ${x}^{2}\u20135x+6=0$

Let f$\left(x\right)$ = ${x}^{2}\u20135x+6$. Now for x = 2, we have $\mathrm{f}\left(2\right)={2}^{2}\u20135\left(2\right)+6=0$

$\therefore $ 2 is a root of ${x}^{2}\u20135x+6=0$

**7.3. Solving a quadratic equation**

We already know how to factorise a quadratic polynomial. We shall now learn how to apply the method of factorisation to solve a quadratic equation.

Basically, there are three methods of finding the roots of a Quadratic Equation.

1. Factorization method

2. Completing Squares

3. Quadratic Formula method

**Factorization method**

Let us consider a quadratic equation,

${x}^{2}$ + 6x + 5 = 0**Step 1: ** Compare the given equation with the general form and identify the terms a, b, and c.

By comparing ${x}^{2}$ + 6x + 5 = 0 with

a${x}^{2}$ + bx + c = 0, we get,

a = 1, b = 6 and c = 5**Step 2:** Multiply the constant with coefficient of ${x}^{2}$ i.e., a × c.

Here a × c = 5 × 1 = 5.**Step 3:** Split the product of a and c into as many factors as possible.

5 = 5 × 1 and 5 = –5 × – 1**Step 4:** Choose those set of factors whose sum is equal to coefficient of x (b).

Here, b = 6 and 5 + 1 = 6.

$\therefore $ The chosen factors are 5 and 1.**Step5:** Split the middle term into two terms whose coefficients are same as that of the chosen factors.

Here, ${x}^{2}$+ 6x + 5 =${x}^{2}$ + 5x + x + 5 = 0**Step 6:** Factorise the quadratic polynomial by equating it to zero and find the values of variable(x).

${x}^{2}$ + 6x + 5 = 0

$\Rightarrow $${x}^{2}$ + 5x + x + 5 = 0

$\Rightarrow $x (x + 5) + 1(x + 5) = 0

$\Rightarrow \left(x+1\right)\left(x+5\right)=0$

Either x + 1 = 0 or x + 5 = 0

$(\because \text{if}a\times b=0\text{then, either}a=0\text{or}b=0)$

i.e., x = –1 or x = –5

$\therefore $ The roots of the quadratic equation are,

x = –1 or x = – 5

**Completing Squares**

Thus far we have discussed the factoring and square root methods for solving quadratic equations algebraically. However, those methods cannot be applied to an equation such as ${x}^{2}$ + 3x – 5 = 0 because the expression ${x}^{2}$ + 3x – 5 is not factorable using integer coefficients and is not in the form appropriate to the square root method (there is a first degree term).

Our interest in this section is to find a method for solving quadratic equations that can be applied to all cases.

If we can take any equation and put it in the form${\left(x+p\right)}^{2}$ = d, where p and d are constants, then all that remains is to apply the square root method – that is, take square roots, solve for x, and simplify the answer. But can all quadratic equations be put in the form ${\left(x+p\right)}^{2}$ = d, where p and d are constants ?

To answer this, we first examine the squares of binomials of the form (x + p), Where p is a constant. When the squares are multiplied out, we call them perfect squares.

First we will square binomials of the form x + p and look at the relationship between the x coefficient and the numerical term. Observe the following :

Coefficient of the x term The numerical term

${\left(x\u20133\right)}^{2}$= ${x}^{2}$ – 6x + 9 – 6 + 9

${\left(x+5\right)}^{2}$ = ${x}^{2}$ + 10x + 25 + 10 + 25

${\left(x+4\right)}^{2}$ = ${x}^{2}$+ 8x + 16 + 8 + 16

Let’s examine what happens when we square (x + p):

${\left(x+p\right)}^{2}$ = ${x}^{2}$ + 2px + ${p}^{2}$

Where,

${x}^{2}$$\to $Square of first term in the binomial.

px$\to $ Twice the Product of the terms in the binomial

${p}^{2}\to $ Square of second term in the binomial.

Given that p is a constant and x is a variable, the middle (first-degree) term coefficient will be 2p and the numerical term will be ${p}^{2}$. What is the relationship between the middle term coefficient, 2p, and the numerical term, ${p}^{2}$ ?

If you take half of 2p and square it, you will get ${p}^{2}$ :

${\left[\frac{1}{2}\left(2\mathrm{p}\right)\right]}^{2}={\left(\frac{2\mathrm{p}}{2}\right)}^{2}={\mathrm{p}}^{2}$

Thus, if you square the binomial (x + p), the square of one-half the middle-term coefficient will yield the numerical term. We know return to our examples.

${\left(x\u20133\right)}^{2}$ = ${x}^{2}$ – 6x + 9.

1. Take half of – 6 to get – 3.

2. Square – 3 to get the numerical term 9.

${\left(x+5\right)}^{2}$ = ${x}^{2}$ + 10x + 25

1. Take half of 10 to get 5.

2. Square 5 to get the numerical term, 25.

Now we will demonstrate how we can take any second-degree equation and put it in the form

${\left(x+p\right)}^{2}$ = d, where p and d are constants. Suppose we have an equation such as ${x}^{2}$ + 6x – 8 = 0

To make it clearer how to make the left side into a perfect square, we first add 8 to both sides of the equation.

${x}^{2}$ + 6x – 8 = 0

Add 8 to both sides of the equation.

${x}^{2}$ + 6x = 8

Now what is missing to make the left side a perfect square? Take half the middle term coefficient, 6, and square it : = ${\left(\frac{1}{2}\times 6\right)}^{2}$ = ${\left(3\right)}^{2}$ = 9

Thus, 9 must be added to make the left side a perfect square. But since we are dealing with equations, we must add 9 to both sides of the equation.

${x}^{2}$ + 6x + 9 = 8 + 9

$\Rightarrow $ ${x}^{2}$+ 6x + 9 = 17

$\Rightarrow $${\left(x+3\right)}^{2}$ = 17

The only difference between the equation

${x}^{2}$ + 6x = 8 and the perfect square version

${x}^{2}$ + 6x + 9 = 17 is that 9 was added to both sides. Why 9? To make the left side a perfect square so that it could be written in factored form. This process of adding a number to make a perfect square is called completing the square.

Now, we can solve this equation by the square root method.

${\left(x+3\right)}^{2}$ = 17

Take square root.

x + 3 = $\pm \sqrt{17}$

Then isolate x by adding – 3 to both sides of the equation.

x = – 3 $\pm \sqrt{17}$

Similarly, any quadratic equation can be put in the form ${\left(x+p\right)}^{2}$ = d, where p and d are constants.

**Quadratic Formula Method**

It is not always convenient to factorise the quadratic equation. Let us derive the quadratic formula which is useful for the solving of all quadratic equations.

**Statement**

If a${x}^{2}$ + bx + c = 0 is a quadratic equation, then the roots are $x=\frac{\u2013b\pm \sqrt{{b}^{2}\u20134ac}}{2a}$**Proof :** Consider the quadratic equation,

a${x}^{2}$ + bx + c = 0, where a,b,c$\in R$ a$\ne $0

a${x}^{2}$ + bx + c = 0

Dividing throughout by a, we get,

${x}^{2}+\frac{b}{a}x+\frac{c}{a}=0$

By adding and subtracting $\frac{{b}^{2}}{4{a}^{2}}$ , we get

${x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}\u2013\frac{{b}^{2}}{4{a}^{2}}+\frac{c}{a}=0$

We can rewrite the equation as

${x}^{2}+2\left(\frac{b}{2a}\right)x+{\left(\frac{b}{2a}\right)}^{2}\u2013{\left(\frac{b}{2a}\right)}^{2}+\frac{c}{a}=0$

We can recognise the first three terms as the formula,

${a}^{2}$ + 2ab + ${b}^{2}$ = ${\left(a+b\right)}^{2}$

$\therefore {\left(x+\frac{b}{2a}\right)}^{2}\u2013{\left(\frac{b}{2a}\right)}^{2}+\frac{c}{a}=0$

$\Rightarrow {\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}\u2013\frac{c}{a}=\left(\frac{{b}^{2}\u20134ac}{4{a}^{2}}\right)$

$\Rightarrow x+\frac{b}{2a}=\pm \frac{\sqrt{{b}^{2}\u20134ac}}{2a}$

$\Rightarrow x=\frac{\u2013b\pm \sqrt{{b}^{2}\u20134ac}}{2a}$

$\Rightarrow x=\frac{\u2013b+\sqrt{{b}^{2}\u20134ac}}{2a}$ and $\frac{\u2013b\u2013\sqrt{{b}^{2}\u20134ac}}{2a}$ are the roots of the quadratic equation .

Generally, roots of quadratic equations are denoted by $\alpha $ and $\beta $ .

Here, $\alpha =\frac{\u2013b+\sqrt{{b}^{2}\u20134ac}}{2a}$ , $\beta =\frac{\u2013b\u2013\sqrt{{b}^{2}\u20134ac}}{2a}$

Here ${b}^{2}$ – 4ac is called discriminant and it is denoted by D or $\u2206$.

**7.4. Roots of Quadratic Equations**

For what values of the variable (x) the quadratic equation gets satisfied are called roots of a quadratic equation.

Let a${x}^{2}$ + bx + c = 0 is a quadratic equation, then the roots are generally denoted by $\alpha and\beta $, where

$\alpha =\frac{\u2013b+\sqrt{{b}^{2}\u20134ac}}{2a}$, $\beta =\frac{\u2013b\u2013\sqrt{{b}^{2}\u20134ac}}{2a}$**Sum and Product of roots**

If $\alpha $and $\beta $ are the two roots of a quadratic equation a${x}^{2}$ + bx + c = 0, then the two roots are , $\alpha =\frac{\u2013b+\sqrt{{b}^{2}\u20134ax}}{2a}$,$\beta =\frac{\u2013b\u2013\sqrt{{b}^{2}\u20134ac}}{2a}$,

we have by addition ,

$\alpha +\beta =\frac{\u2013b+\sqrt{{b}^{2}\u20134ac}\u2013b\u2013\sqrt{{b}^{2}\u20134ac}}{2a}$

$=\u2013\frac{2b}{2a}=\u2013\frac{b}{a}$

$\therefore $ sum of roots $(\alpha +\beta )=\u2013\frac{b}{a}$

and by multiplication, we have

$\alpha \beta =\frac{\left(\u2013b+\sqrt{{b}^{2}\u20134ac}\right)\left(\u2013b\u2013\sqrt{{b}^{2}\u20134ac}\right)}{4{a}^{2}}$

$=\frac{(\u2013b{)}^{2}\u2013\left({b}^{2}\u20134ac\right)}{4{a}^{2}}=\frac{4ac}{4{a}^{2}}=\frac{c}{a}$

$\therefore $ product of roots $\alpha \beta =\frac{c}{a}$

**Writing the Quadratic equation Using its roots**

Let $\alpha $ and $\beta $ are roots of a quadratic equation then the quadratic equation formed (Constructed) by using its roots $\alpha $ and $\beta $ is ${x}^{2}\u2013(\alpha +\beta )x+\alpha \beta =0$

${x}^{2}$– (sum of roots) x + (Product of Roots) = 0 **Example 1:** Let us find the quadratic equation whose roots are 3 and – 2.

The required quadratic equation is = 0

${x}^{2}\u2013(3+(\u20132\left)\right)x+(3x\u20132)=0$

${x}^{2}$ – x – 6 = 0.

**Example 2:** Find the quadratic equation whose roots are $2+\sqrt{3}$and $2\u2013\sqrt{3}$ .

We have sum of roots $\left(\alpha +\beta \right)$= 4, and

Product of roots $\alpha \beta $= 1;

$\therefore $ The required quadratic equation is

${x}^{2}\u2013(\alpha +\beta )x+\alpha \beta =0$

${x}^{2}$ – 4x + 1 = 0 .

**Points to remember**

i) The equation whose roots are $\frac{1}{\alpha},\frac{1}{\beta}$if , $f\left(\frac{1}{x}\right)=0$ if $c\ne 0\text{i.e.},\alpha \beta \ne 0$

ii) The equation whose roots are $\alpha +k,\beta +k$ is

f(x – k) = 0

iii) The equation whose roots are $k\alpha $ and $k\beta $ is $f\left(\frac{x}{k}\right)=0$

iv) The equation whose roots are equal but opposite in sign is f(–x) = 0 i.e., the equation whose roots are $\u2013\alpha $, $\u2013\beta $ is f(–x) = 0

**The Discriminant**

The roots $\alpha and\beta $of the quadratic equation

a${x}^{2}$ + bx + c = 0 are given by $\alpha =\frac{\u2013b+\sqrt{{b}^{2}\u20134ac}}{2a}$,$\beta =\frac{\u2013b\u2013\sqrt{{b}^{2}\u20134ac}}{2a}$ ,

Here, $\u2206={b}^{2}\u20134ac$ and is called Discriminant of quadratic equation.

Then we have the following results:

i) If , $\u2206={b}^{2}\u20134ac=0$ then the roots are real and equal.

ii) If , $\u2206={b}^{2}\u20134ac>0$ then the roots are real and unequal.

iii) If , $\u2206={b}^{2}\u20134ac<0$ then the roots are imaginary.

iv) Given that a, b, c are real. $\u2206<0$ if $\alpha $ and $\beta $ are non-real complex numbers, conjugate to each other.

v) Given that a, b, c are rational numbers, $\u2206$is positive but not a perfect square if $\alpha $ and $\beta $ are conjugate surds like $c+\sqrt{d}$ and $c\u2013\sqrt{d}$ .

By applying these tests the nature of the roots of any quadratic equation can be determined without actually finding the roots.

**Important Properties of Roots**

i) If the roots are equal in magnitude but opposite in sign, then coefficient of x = 0 i.e., b = 0.**Example:** ${x}^{2}$ – 4 = 0 $\Rightarrow x=\pm 2$

clearly, b = 0

ii) If unity is a root of a${x}^{2}$ + bx + c = 0 then the other root is c/a.**Example: **${x}^{2}$ – 4x + 3 = 0 $\Rightarrow $x = 1, x = 3

$\therefore $ Other root c/a = 3/1 = 3

iii) Irrational roots of a quadratic equation with rational coefficients occur in conjugate pairs. If $p+\sqrt{q}$ is a root of a${x}^{2}$ + bx + c = 0, then $p\u2013\sqrt{q}$ is also a root.**Example:** $2+\sqrt{3},2\u2013\sqrt{3}$ are the roots of ${x}^{2}$ – 4x + 1 = 0

iv) Imaginary or complex roots of a quadratic equation with real coefficients occur in conjugate pairs. If p + iq is a root of

a${x}^{2}$ + bx + c, then p – iq is also a root of the equation.

Where $i=\sqrt{\u20131}$**Example:** 3 + 5i, 3 – 5i are the roots of ${x}^{2}$ – 6x + 34 = 0

**Common roots of quadratic equations**

i) If both roots are common to two equations

${a}_{1}$${x}^{2}$ + ${b}_{1}$x + ${c}_{1}$ = 0 and ${a}_{2}$${x}^{2}$ + ${b}_{2}$x + ${c}_{2}$ = 0, [where ${a}_{1}$, ${a}_{2}$$\ne $0 and (${a}_{1}$${b}_{2}$ – ${a}_{2}$${b}_{1}$)$\ne $0]

then, $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$

ii) If only one root is common to two equations

${a}_{1}$${x}^{2}$ + ${b}_{1}$x + ${c}_{1}$ = 0 and ${a}_{2}$${x}^{2}$ + ${b}_{2}$x + ${c}_{2}$ = 0, [where ${a}_{1}$, ${a}_{2}$$\ne $ 0

and (${a}_{1}$${b}_{2}$ – ${a}_{2}$${b}_{1}$)$\ne $0]

then ${\left({c}_{1}{a}_{2}\u2013{c}_{2}{a}_{1}\right)}^{2}$ = (${b}_{1}$${c}_{2}$ – ${b}_{2}$${c}_{1}$)(${a}_{1}$${b}_{2}$ – ${a}_{2}$${b}_{1}$) and the

common root =$\frac{{c}_{1}{a}_{2}\u2013{c}_{2}{a}_{1}}{{a}_{1}{b}_{2}\u2013{a}_{2}{b}_{1}}$