NCERT solutions for class 10 maths Chapter 4 Quadratic equations 4 Nature of roots | SureSolv

You are here

For guided learning and practice on Number system and HCF LCM, follow Comprehensive Suresolv Number system HCF LCM Guide with all articles

NCERT solutions for class 10 maths Chapter 4 Quadratic equations 4 Nature of roots

Nature of roots of a quadratic equation

NCERT-solutions-class-10-maths-ch-4-pt-4

This is the fourth and last part of the series on Quadratic equations in Chapter 4.

It complements the section with Exercise 4 of Chapter 4 of NCERT Class 10 Maths, in short, Ex 4.4 class 10.

In this session we will cover,

  1. What are the nature of roots of a quadratic equation,
  2. What is the discriminant of a quadratic equation, and,
  3. How to determine whether the roots of a quadratic equation are real, imaginary or distinct using the quadratic equation formula by Sreedhar Acharya.

Discussions are divided into three sections:

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

To Skip the Concept Explanations and Go straight to the Solutions, click here.

What are the nature of roots of a quadratic equation and how to determine the nature by the discriminant of the equation

We already know that the roots of a general quadratic equation $ax^2+bx+c=0$ are given by the standard quadratic equation formula,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

In this formula, the key component is $b^2-4ac$ in the numerator. This is the discriminant of the quadratic equation.

As $b^2-4ac$ is under square root,

if it is 0, that is, $b^2-4ac=0$, then both the root values are equal. We would say, the equation does not have two distinct roots. In this case, the single value of the two roots is obviously real.

if the discriminant is positive, that is, $b^2 - 4ac \gt 0$, the result of the square root will be real, and we will have both the roots real and distinct.

In the third unusual case,

when the discriminant is negative, that is, $b^2 -4ac \lt 0$, taking square root of the negative result would lead to imaginary value. We would say, the quadratic equation does not have any real roots.

These are three main possibilities.

From the quadratic equation formula, you can also determine whether the roots are integers, rational fractions, or irrational involving surd,

If the discriminant $b^2-4ac$ is a perfect square, you will have the roots either rational fractions or integers, but not irrational surd.

But if the discriminant $b^2-4ac$ is not a perfect square, the roots will involve surd.

Finally, when $b^2-4ac$ is a perfect square and both values of the numerator are multiples of denominator $2a$, the roots will be real and integers.

Let us apply these concepts on a few example problems.

Problem example 1.

Find whether the roots of the following quadratic equation are real or not.

$3x^2-4x+2=0$.

Solution to problem example 1.

Here, $a=3$, $b=-4$ and $c=2$.

So the discriminant value is,

$b^2-4ac=(-4^2)-4\times{3}\times{2}=16-24=-8 \lt 0$.

The equation does not have any real roots.

Problem example 2.

There is a circular park with two diametrically opposite gates on the boundary separated by a distance of 10 meters. A lamp post is to be erected on the boundary in such a way that the difference of its distances from the gates would be 3 meters. Is it possible to erect such a lamp post?

Solution to example problem 2.

Our first job is to form the quadratic equation representing the descriptive problem.

By the property of a circle, a diameter will form an angle of $90^0$ at any point on the periphery of the circle. So the line joining the two gates being the diameter of the circular garden, it will be the 10 meter long hypotenuse of a right-angled triangle at the point where the lamp post is to be erected on the boundary.

The following picture shows the situation.

ncert-class-10-maths-ch-4-part-4-pic1.png

Let shorter side of this right angled triangle be $x$ meters. So the longer side will be $(x+3)$ meters long.

By Pythagoras theorem, sum of squares of two sides is equal to the square of the hypotenuse in a right-angled triangle. So,

$x^2+(x+3)^2=10^2$,

Or, $2x^2+6x-91=0$.

If the roots of this equation are real we can declare that erection of the lamp post satisfying the conditions will be possible.

Here, $a=2$, $b=6$ and $c=-91$.

The value of the discriminant is,

$b^2-4ac=6^2+4\times{2}\times{91}=764$.

As the discriminant is positive, the roots are real, and the positive real roots will be the values of $x$.

It will be possible to erect the lamp post.

The value of the roots are,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$,

Or, $x=\displaystyle\frac{-6 \pm \sqrt{4\times{191}}}{4}$

$=\displaystyle\frac{-6 \pm 2\sqrt{191}}{4}$

$=\displaystyle\frac{-3 \pm \sqrt{191}}{2}$.

The negative value for length being invalid, the positive value of $x$, the length of the shorter side will be,

$x=\displaystyle\frac{-3 + \sqrt{191}}{2}$.

And the longer side being more than the shorter side by 3 meters, it will be,

$(x+3)=3+\displaystyle\frac{-3 + \sqrt{191}}{2}=\displaystyle\frac{3 + \sqrt{191}}{2}$.

We'll solve now the exercise problems.

This is the third and last component of the session.

Solved problems: NCERT Class 10 maths chapter 4 exercise 4: Nature of roots of a quadratic equation

Problem 1.i.

Find the nature of the roots of the following quadratic equations. If real roots exist, find them.

$2x^2-3x+5=0$.

Solution to problem 1.i.

The coefficients are,

$a=2$, $b=-3$, $c=5$.

The discriminant is,

$b^2-4ac=9-40=-31$.

As the discriminant under square root is negative, no real roots exist for this equation.

Answer: No real roots exist.

Problem 1.ii.

Find the nature of the roots of the following quadratic equations. If real roots exist, find them.

$3x^2-4\sqrt{3}x+4=0$.

Solution to problem 1.ii.

The coefficients are,

$a=3$, $b=-4\sqrt{3}$, $c=4$.

The discriminant is,

$b^2-4ac=48-48=0$.

The roots are given by,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$,

Or, $x=\displaystyle\frac{4\sqrt{3} \pm 0}{6}=\displaystyle\frac{2}{\sqrt{3}}$.

Both the roots are real and of equal value, $x=\displaystyle\frac{2}{\sqrt{3}}$.

Answer: Both roots are real and of equal value, $x\displaystyle\frac{2}{\sqrt{3}}$.

Problem 1.iii.

Find the nature of the roots of the following quadratic equations. If real roots exist, find them.

$2x^2-6x+3=0$.

Solution to problem 1.iii.

The coefficients are,

$a=2$, $b=-6$, $c=3$.

The discriminant is,

$b^2-4ac=36-24=12$.

The discriminant being positive, both the roots of the equation are real and distinct.

The roots are,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$,

Or, $x=\displaystyle\frac{6 \pm 2\sqrt{3}}{4}=\frac{3 \pm \sqrt{3}}{2}$,

So the roots are,

$x=\displaystyle\frac{\sqrt{3}}{2}(\sqrt{3}+1)$, and

$x=\displaystyle\frac{\sqrt{3}}{2}(\sqrt{3}-1)$.

Answer: Roots are real distinct, $\displaystyle\frac{\sqrt{3}}{2}(\sqrt{3}+1)$, and $\displaystyle\frac{\sqrt{3}}{2}(\sqrt{3}-1)$.

Problem 2.i.

Find the values of $k$ for the following quadratic equation, so that it has two equal roots.

$2x^2+kx+3=0$.

Solution to problem 2.i.

The value of the discriminant is to be zero for the two roots to be equal.

In this case, the coefficients are,

$a=2$, $b=k$, and $c=3$.

Value of the discriminant is,

$b^2-4ac=k^2-24=0$, for the roots to be equal.

So, $k^2-24=0$,

Or, $k^2=24$,

Or, $k=\pm 2\sqrt{6}$.

Answer: $k=\pm 2\sqrt{6}$.

Problem 2.ii.

Find the values of $k$ for the following quadratic equation, so that it has two equal roots.

$kx(x-2)+6=0$.

Solution to problem 2.ii.

The value of the discriminant is to be zero for the two roots to be equal.

To find the coefficient values, you have to first simplify the given equation in the standard form of, $a^2+bx+c=0$, where $a$, $b$ and $c$ are the coefficients.

The given equation is,

$kx(x-2)+6=0$,

Or, $kx^2-2kx+6=0$

So in this case the values of the coefficients are,

$a=k$, $b=-2k$, and $c=6$.

Value of the discriminant is,

$b^2-4ac=4k^2-24k=0$, for the two roots to be of equal value.

Or, $k^2-6k=0$,

Or, $k(k-6)=0$.

The two values of $k$ satisfying this second quadratic equation are, $k=0$ and $k=6$.

But if $k=0$, the given quadratic equation turns out to be invalid.

So, $k=6$ for the two roots to be of equal value.

Answer: $k=6$.

Problem 3.

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is, 800 sq m? If so, find its length and breadth.

Solution to problem 3.

First form the quadratic equation in a single variable.

Let us assume the breadth of the mango grove to be $x$ m. So its length is $2x$ m and area is,

$x\times{2x}=800$,

Or, $2x^2=800$,

Or, $x^2=400$,

Or, $x=\pm 20$.

As breadth cannot be negative, its value is, $x=20$ m and the length is $40$ m. It is possible to design a rectangular mango grove with these dimensions that satisfy the problem conditions.

Answer: Possible. Breadth 20 m, length 40 m.

Problem 4.

Is the following situation possible? If so, determine their present ages.

The sum of ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution to problem 4.

Let present age of one friend be $x$ years. So the sum of their ages being 20 years, the age of the second friend is $(20-x)$ years.

Four years ago their respective ages were,

for first friend: $(x-4)$ years, and,

for second friend: $(20-x-4)=(16-x)$ years.

The product of these two is,

$(x-4)(16-x)=48$, as per the problem,

Or, $-x^2+20x-64=48$,

Or, $x^2-20x+112=0$.

The coefficients are,

$a=1$, $b=-20$ and $c=112$.

The value of the discriminant is,

$b^2-4ac=400-448=-48$.

The discriminant being negative, the given situation is not possible. The quadratic equation does not have any real roots.

Answer: Not possible.

Problem 5.

Is it possible to design a rectangular park of perimeter 80 m and area 400 sq m? If so, find its length and breadth.

Solution to problem 5.

Let us assume the breadth of the park is $x$ m and length $y$ m. Its perimeter is then,

$2(x+y)=80$,

Or, $x+y=40$,

Or, $y=40-x$.

Also the area is product of the length and breadth,

$x(40-x)=400$,

Or, $-x^2+40x=400$,

Or, $x^2-40x+400=0$.

The coefficients are,

$a=1$, $b=-40$ and $c=400$.

The discriminant is,

$b^2-4ac=1600-1600=0$.

Both the roots are then of equal real value.

The root value is,

Breadth $x=\displaystyle\frac{40}{2}=20$ m, and,

Length $y=40-x=20$ m.

Answer: Yes, it is possible. Both length and breadth are of equal value 20 m.

This ends Chapter 4 on Quadratic equations.


NCERT Solutions for Class 10 Maths

Chapter 1: Real Numbers

NCERT Solutions for Class 10 Maths on Real numbers part 3, HCF and LCM by factorization and problem solutions

NCERT Solutions for Class 10 Maths on Real numbers part 2, Euclid’s division algorithms, HCF and problem solutions

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

Chapter 2: Polynomials

NCERT Solutions for Class 10 Maths Chaper 2 Polynomials 1 Geometrical Meaning of Zeroes of 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 5 Algebraic solution by Cross Multiplication

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

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 4 Nature of roots of a quadratic equation

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 2 Ratio values for selected angles

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 3 Complementary angle relations

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


Share