## Introduction to quadratic equations

This is the first of the four part series on **Quadratic equations**. It complements Chapter 4 of NCERT Class 10 Maths. In this introductory session two aspects quadratic equations are explored.

- What are quadratic equations and,
- How to check that an equation is quadratic or not.

Discussions are divided into **three sections:**

**First:**concepts explained in an easy to understand way,**Second:**example problems used for establishing the concepts clearly and,**Third:**the solutions to the problems in the 1st exercise of NCERT Chapter 4.

### Algebraic expressions and quadratic equations

One of the most important topics in Maths, Algebra, starts with the concept of representing real world attributes such as length, breadth, area, price, time, distance, number of people and so on, by symbols.

This is what we call—ABSTRACTION.

A simple example of this concept is—the case when you want to find the length and breadth of a rectangular field with its area given.

#### Problem description stage 1

Find the length and breadth of a rectangular field of area 495 sq m.

The two independent attributes of the rectangular field are its length and breadth.

As a first step in defining the situation, algebraic representation starts with using a symbol, say L, for length and a second symbol, say B, for breadth of the field.

If you use now a third symbol, say A, to represent the area of the field, it becomes very easy for you to express the area as the product of length and breadth in a concise form as,

$L\times{B}=A$

The advantages are obvious—

- You don’t have to be confined to describing the problem in natural language, and,
- You can mathematically deal with this simple model representing the real world problem.

L, B, and A are the three variables, value of only one of which is known and you have to find the values of the other two. The relationship between the three attributes represented by the three variables is mathematically modeled in the equation above.

As value of area A is known, Number of variables is reduced to 2, with 495 as a Constant of fixed value that does not change.

The two symbols L and B are of,

- same unit metres, and,
- unknown value.

These are the two Primary and Essential requirements for representing attributes of the rectangular field as VARIABLES so that you can use the equation for finding the values of each variable mathematically.

You have not yet reached the stage where you can apply mathematical METHODS to find the values of the variables.

This is because—it is a mathematical truth that you cannot find the values of variables by solving one or more than one equation if number of independent equations is less than the number of variables.

In our problem, you have one equation, but two variables in the equation—you need more information.

So to make the problem solvable, more information is supplied by modifying it.

#### Problem description stage 2

Find the length and breadth of a rectangular field of area 495 sq m where length is 3 more than twice the breadth.

You can represent this new information mathematically as,

$L=2B+3$.

So the original equation is changed to,

$LB=495$,

Or, $(2B+3)B=495$,

Or, $2B^2+3B-495=0$.

Now this is a single equation in one variable B.

This is what is called as a **Quadratic equation.**

The **LHS of the equation is a polynomial of degree 2** (highest power of variable B is 2) **equated to 0.**

These are the **two essential requirements for a quadratic equation.**

As general conventions, the symbols $x$, $y$, $z$ and so on are used as variables in dealing with mathematical relations such as this quadratic equation.

To fall in line with the general practice, we will replace symbol B for breadth by symbol $x$. The quadratic equation is expressed then as,

$2x^2+3x-495=0$. $\qquad \qquad \qquad..........................(1)$

We have to solve this equation.

#### Standard form of a quadratic equation

The standard form of a quadratic equation is,

$ax^2+bx+c=0$.

$x$ is the single variable having three terms in the LHS with power reducing from 2 to 1 to 0. The third term is thus a constant.

#### Trivial form of quadratic equations

**Case 1**

If the constant term $c=0$, our equation (1) is reduced to,

$2x^2 +3x=0$,

Or, $x(2x+3)=0$.

The two possible values of $x$ are, $x=0$, or $x=-\frac{3}{2}$.

It is a trivial problem where finding the solution is too easy.

**Case 2**

If the coefficient $b$ of $x$ is 0, the equation (1) reduces to,

$2x^2-495=0$,

Or, $x^2=\displaystyle\frac{495}{2}$,

Or, $x=\pm3\sqrt{\displaystyle\frac{55}{2}}$.

Again it is too easy to find the solution.

#### Solving a quadratic equation to find its two roots

Our quadratic equation to solve is,

$2x^2+3x-495=0$. $\qquad \qquad \qquad..........................(1)$

As the highest power of the variable $x$ in the equation is 2, $x$ will have 2 values that satisfy the equation. These are called the roots of the quadratic equation.

The roots may be real numbers or imaginary numbers, may be positive or negative real numbers—there is no constraint on the values of the roots of a quadratic equation except that the values will be defined (not infinite).

Question now is how you would get the values of the roots of the equation (1).

Assuming that you know how to factorize the quadratic expression in LHS, the equation can be expressed as,

$2x^2+3x-495=0$,

Or, $(2x+33)(x-15 )=0$$\qquad \qquad \qquad......................(2)$

To satisfy the equation either the first or the second factor in LHS must be 0, that is,

$2x+33=0$, which gives, $x=-\displaystyle\frac{33}{2}$,

Or, $x-15=0$, which gives, $x=15$.

These two possible values of $x$ are the two roots of the quadratic equation.

But as breadth $x$ cannot be negative, in our real world problem, $x\neq -\displaystyle\frac{33}{2}$.

So breadth $B=x=15$ metres, and

Length $L=2B+3=2x+3=33$ metres.

In this session we won't cover how to solve a quadratic equation. Instead we'll deal with two introductory aspects of quadratic equations,

- How to form a quadratic equation from a descriptive problem, and,
- How to check whether an equation is quadratic or not.

### How to form a quadratic equation from a descriptive problem

We' will use a few example problems to learn how to form a quadratic equation from description of a problem.

#### Problem example 1.

Represent the following description mathematically.

Out of two positive integers, the larger integer is 7 more than five times the smaller one. The product of the two integers is 1938. We need to find the value of the integers.

#### Solution example 1.

Let the smaller integer be $x$. So the larger integer is,

$\text{larger integer}=5x+7$.

The product of the two is,

$x(5x+7)=1938$

Or, $5x^2+7x-1938=0$.

This conforms to the standard form of quadratic equations, $ax^2+bx+c=0$ with $a=5$, $b=7$ and $c=-1938$.

**Note:** To find the values of the integers, you have to find first the roots of the quadratic equation. By factorization the quadratic equation is expressed as,

$(5x+102)(x-19)=0$.

It follows,

Either $5x+102=0$ leading to $x=-\displaystyle\frac{102}{5}$, an invalid value,

Or, $x-19=0$ leading to $x=19$, the valid value of the smaller integer.

Larger integer is, $5\times{19}+7=102$.

#### Problem example 2.

A train travels a distance of 660 km at a uniform speed. If the speed were 6 km/hr more, the train would have taken 1 hour less to cover the same distance of 660 km. We need to find the original speed of the train.

#### Solution example 2.

Speed time distance relationship is given by,

$\text{time}=\displaystyle\frac{\text{distance}}{\text{speed}}$

Let us assume, $t$ as the original time taken to cover 660 km at a speed of $s$ km/hr.

So originally,

$t=\displaystyle\frac{660}{s}$.$\qquad \qquad \qquad \qquad...........................(3)$

The changed speed is, $(s+6)$ km/hr while the changed time taken to cover 660 km at this changed speed is $(t-1)$ hr. The speed time distance relationship in this case is,

$t-1=\displaystyle\frac{660}{s+6}$.$\qquad \qquad \qquad..........................(4)$

To eliminate the second variable of time $t$, subtract equation (4) from equation (3). Result is,

$\displaystyle\frac{660}{s}-\displaystyle\frac{660}{s+6}=1$,

Or, $660(s+6-s)=s(s+6)$,

Or, $s^2+6s-3960=0$.

This conforms to the standard form of quadratic equations, $ax^2+bx+c=0$ with $a=1$, $b=6$ and $c=-3960$, and $x$ stands for $s$.

**Note:** To find the values of the integers, you have to find first the roots of the quadratic equation. By factorization the quadratic equation is expressed as,

$(s+66)(s-60)=0$.

It follows,

Either $s+66=0$ leading to $s=-66$, an invalid value,

Or, $s-60=0$ leading to $s=60$ km/hr, the original speed.

### How to check whether an equation is quadratic or not

We would explain by using a few problem examples.

#### Problem example 3.

Check whether the following is a quadratic equation.

$x^2-5x=(x+3)(x+11)$

#### Solution example 3.

Given equation is,

$x^2-5x=(x+3)(x+11)$

Or, $x^2-5x=x^2+14x+33$,

Or, $19x+33=0$.

In this equation, coefficient of $x^2$ is 0, and so the equation is linear, not quadratic.

**Technique used is**—

Expand all the products to individual terms and bring all the non-zero terms on the LHS. Then arrange the terms in decreasing power of $x$ from left to right.

In this form, it takes just a few seconds to identify the nature of the equation.

#### Problem example 4.

Check whether the following is a quadratic equation.

$(2x+7)(3x-8)=(x-1)(x-13)$.

#### Solution example 4.

Given equation is,

$(2x+7)(3x-8)=(x-1)(x-13)$

Or, $6x^2+5x-56=x^2-14x+13$,

Or, $5x^2+19x-69=0$.

This is a quadratic equation in standard form of, $ax^2+bx+c=0$ with $a=5$, $b=19$, and $c=-69$.

#### Problem example 5.

Check whether the following is a quadratic equation.

$(x-1)^3=x^3-4x^2+x+4$.

#### Solution example 5.

Given equation is,

$(x-1)^3=x^3-4x^2+x+4$,

Or, $x^3-3x^2+3x-1=x^3-4x^2+x+4$,

Or, $x^2+2x-5=0$.

This is a quadratic equation in standard form of, $ax^2+bx+c=0$, where $a=1$, $b=2$, and $c=-5$.

We'll now take up solving the exercise problems.

### Problems to solve: Exercise 1 Chapter 4 NCERT solutions for Class 10 maths: Quadratic equations

#### Problem 1.i.

Check whether the following is a quadratic equation.

$(x+1)^2=2(x-3)$.

#### Solution to Problem 1.i

Given equation is,

$(x+1)^2=2(x-3)$,

Or, $x^2+2x+1=2x-6$,

Or, $x^2+7=0$,

Or, $x=\pm\sqrt{-7}$.

The equation is a trivial form of quadratic equation with imaginary values of the roots of $x$.

**Answer:** Yes.

#### Problem 1.ii.

Check whether the following is a quadratic equation.

$x^2-2x=(-2)(3-x)$.

#### Solution to problem 1.ii.

Given equation is,

$x^2-2x=(-2)(3-x)$,

Or, $x^2-2x=2x-6$,

Or, $x^2-4x+6=0$.

This a quadratic equation of standard form $ax^2+bx+c=0$ with $a=1$, $b=-4$, and $c=6$.

**Answer:** Yes.

#### Problem 1.iii.

Check whether the following is a quadratic equation.

$(x-2)(x+1)=(x-1)(x+3)$.

#### Solution to problem 1.iii.

$(x-2)(x+1)=(x-1)(x+3)$,

Or, $x^2-x-2=x^2+2x-3$,

Or, $3x-1=0$.

This is a linear equation with highest power in $x$ as 1.

**Answer:** No.

#### Problem 1.iv.

Check whether the following is a quadratic equation.

$(x-3)(2x+1)=x(x+5)$.

#### Solution to problem 1.iv.

Given equation is,

$(x-3)(2x+1)=x(x+5)$,

Or, $2x^2-5x-3=x^2+5x$,

Or, $x^2-10x-3=0$.

This is a quadratic equation of standard form $ax^2+bx+c=0$, with $a=1$, $b=-10$, and $c=-3$.

**Answer:** Yes.

#### Problem 1.v.

Check whether the following is a quadratic equation.

$(2x-1)(x-3)=(x+5)(x-1)$.

#### Solution to problem 1.v.

Given equation is,

$(2x-1)(x-3)=(x+5)(x-1)$,

Or, $2x^2-7x+3=x^2+4x-5$,

Or, $x^2-11x+8=0$.

This is a quadratic equation of standard form $ax^2+bx+c=0$, with $a=14$, $b=-11$ and $c=8$.

**Answer:** Yes.

#### Problem 1.vi.

Check whether the following is a quadratic equation.

$x^2+3x+1=(x-2)^2$.

#### Solution to problem 1.vi.

Given problem is,

$x^2+3x+1=(x-2)^2$,

Or, $x^2+3x+1=x^2-4x+4$,

Or, $7x-3=0$.

This is a linear equation with highest power in $x$ as 1.

**Answer:** No.

#### Problem 1.vii.

Check whether the following is a quadratic equation.

$(x+2)^3=2x(x^2-1)$.

#### Solution to problem 1.vii.

Given equation is,

$(x+2)^3=2x(x^2-1)$,

Or, $x^3+6x^2+12x+8=2x^3-2x$,

Or, $x^3-6x^2-14x-8=0$.

Highest power of $x$ is 3 in this equation. So it not a quadratic equation.

**Answer:** No.

#### Problem 1.viii.

Check whether the following is a quadratic equation.

$x^3-4x^2-x+1=(x-2)^3$.

#### Solution to problem 1.viii.

Given equation is,

$(x^3-4x^2-x+1)=(x-2)^3$,

Or, $x^3-4x^2-x+1=x^3-6x^2+12x-8$,

Or, $2x^2+11x+9=0$.

This is a quadratic equation of the standard form $ax^2+bx+c=0$, with $a=2$, $b=11$ and $c=9$.

**Answer:** Yes.

#### Problem 2.i.

Represent the following situation in the form of a quadratic equation.

The area of a rectangular plot is 528 sq m. The length of the plot (in metres) is 1 more than twice its breadth. We need to find the length and breadth of the plot.

#### Solution to Problem 2.i.

Let the breadth of the plot be $x$ m.

So its length is, $(2x+1)$ m.

And area is,

$x(2x+1)=528$,

Or, $2x^2+x-528=0$, where $x$ is the breadth (in metres) of the plot.

This is the quadratic equation representing the description of the problem.

This equation is of standard form of $ax^2+bx+c=0$ with $a=2$, $b=1$, and $c=-528$.

**Note: **Factorizing the equation you would get,

$(2x+33)(x-16)=0$.

Valid value of breadth $x=16$ m, and length $2x+1=33$ m.

#### Problem 2.ii.

Represent the following situation in the form of a quadratic equation.

The product of two consecutive positive integers is 306. We need to find the integers.

#### Solution to problem 2.ii.

Let the smaller integer be $x$. So the larger integer is $(x+1)$.

Product of the two integers is,

$x(x+1)=306$,

Or, $x^2+x-306=0$, where $x$ is the value of the smaller integer.

This is the quadratic equation representing the description of the problem.

This equation is of standard form of $ax^2+bx+c=0$ with $a=1$, $b=1$, and $c=-306$.

**Note:** Factorizing the LHS, you would get,

$(x+18)(x-17)=0$.

Valid value of $x$, the smaller integer is 17 derived from $(x-17)=0$. So, the larger integer is 18.

The other value, $x=-18$ is invalid.

#### Problem 2.iii.

Represent the following situation in the form of a quadratic equation.

Rohan's mother is 26 years older than him and product of their ages (in years) 3 years from now will be 360. We would like to find Rohan's present age.

#### Solution to problem 2.iii.

Let Rohan's present age be $x$ years.

So his mother's present is, $(x+26)$ years.

After three years Rohan's age will be, $(x+3)$ years and his mother's age will be $(x+29)$ years.

Product of their ages after 3 years will be,

$(x+3)(x+29)=360$,

Or, $x^2+32x-273=0$, where $x$ in years is Rohan's present age.

This is the quadratic equation representing the description of the problem.

This equation is of standard form of $ax^2+bx+c=0$ with $a=1$, $b=32$, and $c=-273$.

**Note:** Factorizing the LHS of the equation you would get,

$x^2+32x-273=(x+39)(x-7)=0$.

With $(x+39)=0$, $x=-39$ an invalid value.

With $(x-7)=0$, $x=7$, the present age of Rohan.

#### Problem 2.iv.

Represent the following situation in the form of a quadratic equation.

A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/hr less, then it would have taken 3 hours more to cover the same distance. We need to find the original speed of the train (in km/hr).

#### Solution to problem 2.iv.

Let the original speed of the train be $s$ km/hr, and time taken to cover the distance of 480 km be $t$ hr.

Speed time distance relationship is,

$\text{time}=\displaystyle\frac{\text{distance}}{\text{speed}}$.

Before the decrease in speed then,

$t=\displaystyle\frac{480}{s}$$\qquad \qquad \qquad \qquad...........................(5)$

When speed decreased by 8 km/hr, the changed speed became, $(s-8)$ km/hr.

At this speed time to cover the distance increased by 3 hrs. The changed time to cover became, $(t+3)$ hr.

The relationship between the speed and time after decrease in speed is then,

$t+3=\displaystyle\frac{480}{s-8}$$\qquad \qquad \qquad..........................(6)$

Eliminate time variable $t$ by subtracting equation (5) from equation (6). Result is,

$3=\displaystyle\frac{480}{s-8}-\displaystyle\frac{480}{s}$,

Or, $3s(s-8)=480(s-s+8)=3840$,

Or, $3s^2-24s-3840=0$,

Or, $s^2-8s-1280=0$, eliminating the common factor of 3 from the three terms.

This is the quadratic equation representing the description of the problem where $s$ in km/hr is the original speed of the train.

This equation is of standard form of $ax^2+bx+c=0$ with $a=1$, $b=-8$, and $c=-1280$.

**Note:** Factorizing the LHS of the equation you would get,

$s^2-8s-1280=(s-40)(s+32)=0$.

The value of root $s=-32$, from $(s+32)=0$ is invalid and so the valid value of the original speed is obtained as,

$(s-40)=0$,

Or, $s=40$ km/hr.

### NCERT Solutions for Class 10 Maths

#### Chapter 1: Real Numbers

**NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions**

#### Chapter 2: Polynomials

#### Chapter 3: Linear Equations

**NCERT solutions for class 10 maths Chapter 3 Linear equations 7 Problem Collection**

**NCERT solutions for class 10 maths Chapter 3 Linear equations 6 Reducing non-linear to linear form**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 4 Algebraic solution by Elimination**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 3 Algebraic solution by Substitution**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 2 Graphical solutions**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation.**

#### Chapter 4: Quadratic equations

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 1 What are quadratic equations**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 2 Solving by factorization**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 3 Solution by Completing the square**

#### Chapter 6: Triangles

**NCERT solutions for class 10 maths chapter 6 Triangles 1 Similarity of Triangles and Polygons**

**Solutions to Exercise 2 Chapter 6 NCERT X Maths, Characteristics of Similar triangles**

#### Chapter 8: Introduction to Trigonometry, Concepts and solutions to exercise problems

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 1 Trigonometric Ratios*

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 4 Trigonometric identities*

#### Chapter 8: Introduction to Trigonometry, only solutions to selected problems

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1**