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Quadratic Equations

Complete Class 10 chapter notes with Indian context, solved examples, Shreedharacharya's formula, discriminant logic, and exam-ready application questions.

Chapter Roadmap

01What Is a Quadratic Equation?02Roots of a Quadratic Equation 03Factorisation Method 04Completing the Square 05Quadratic Formula 06Discriminant and Nature of Roots 07Word Problems and Applications

What Is a Quadratic Equation?

A quadratic equation has degree 2 and can always be written in standard form.

Indian Analogy

Think of a cricket ball hit high into the air. Its path bends instead of moving like a straight line. That curved path is modelled by a quadratic expression.

Identification Checklist

Simplify the equation fully first.
Check the highest power of the variable.
If the highest power is 2 and the coefficient is non-zero, it is quadratic.

Not Quadratic

$x^3-x^2-3=0$ because the highest power is 3
$x^2+\frac{1}{x^2}=2$ because it is not a polynomial equation
$x+2\sqrt{x}-3=0$ because the variable appears under a root

A quadratic equation in one variable is an equation that can be written in the form

ax^2+bx+c=0

, where

a

b

, and

c

are real numbers and

a\ne0

If the coefficient of

x^2

becomes zero, the equation is no longer quadratic. That is why

a\ne0

is not a small condition but the main condition.

Quadratic equations appear naturally in projectile motion, area problems, speed problems, age relations, and product-type number problems.

ax^2+bx+c=0,\quad a\ne0

Standard form of a quadratic equation.

Solved Example 1

Identify a quadratic equation

Which of the following are quadratic equations?

(i)

x^2+\dfrac{3}{x}=x^2

(ii)

x^2+\dfrac{1}{x^2}=2

(iii)

3x^2-4x+2=2x^2-2x+4

Show solution

(i) simplifies to $\dfrac{3}{x}=0$ , so it is not quadratic.

(ii) becomes $x^4-2x^2+1=0$ , which is degree 4.

(iii) simplifies to $x^2-2x-2=0$ , which is quadratic.

So only (iii) is quadratic.

Solved Example 2

Tray area problem

The length of a rectangular tray is 1 more than twice its breadth. Its area is

528\text{ cm}^2

. Form the quadratic equation.

Show solution

Let breadth be $x$ cm. Then length is $2x+1$ cm.

Area gives $x(2x+1)=528$ .

So the quadratic equation is

2x^2+x-528=0

Practice

Check which of these are quadratic:

x^2-6x+4=0

x^3-4x+1=0

3x^2-4x+2=2x^2-2x+4

, and

x+\dfrac{3}{x}=x^2

Roots of a Quadratic Equation

A root is a value that makes the quadratic expression equal to zero.

Arrow Analogy

If an arrow follows a parabolic path, its roots tell where the path touches the ground. That is why quadratics can have two such ground-touching points.

If substituting a number into the left-hand side gives 0, that number is a root of the equation.

A quadratic equation can have at most two roots. These roots may be distinct or equal.

f(\alpha)=0

Then

\alpha

is a root of the equation

f(x)=0

Solved Example 3

Check whether a value is a root

x=\dfrac{1}{2}

a root of

2x^2-7x+3=0

Show solution

Substitute $x=\dfrac{1}{2}$ :

2\left(\frac{1}{2}\right)^2-7\left(\frac{1}{2}\right)+3=\frac{1}{2}-\frac{7}{2}+3=0

So $x=\dfrac{1}{2}$ is a root.

Solved Example 4

Find an unknown coefficient from a root

x=2

is a root of

2x^2+kx-6=0

, find

k

and the other root.

Show solution

Since $x=2$ is a root,

2(2)^2+2k-6=0 \Rightarrow 8+2k-6=0 \Rightarrow k=-1

The equation becomes $2x^2-x-6=0$ .

Factorise: $(x-2)(2x+3)=0$ .

So the other root is $x=-\dfrac{3}{2}$ .

Practice

1. Is

x=3

a root of

x^2-5x+6=0

?
2. Find

k

x=-1

is a root of

x^2+kx+2=0

Factorisation Method

Split the middle term, group, and apply the zero-product rule.

Middle-Term Split

For $ax^2+bx+c$ , find two numbers whose product is $ac$ and whose sum is $b$ .

Indian Analogy

It is like splitting a big box of sweets into two smaller neat boxes. Once the quadratic breaks cleanly, the roots become easy to read.

Factorisation is the quickest method when the expression breaks into two linear factors neatly.

The standard trick is to find two numbers whose product is

ac

and whose sum is

b

(px+q)(rx+s)=0

Then either

px+q=0

rx+s=0

(x-\alpha)(x-\beta)=0

So the roots are

x=\alpha

and

x=\beta

Solved Example 5

Simple factorisation

Solve

x^2+6x+5=0

by factorisation.

Show solution

Product = 5 and sum = 6, so use 1 and 5.

x^2+x+5x+5=0 \Rightarrow x(x+1)+5(x+1)=0

(x+1)(x+5)=0

Roots: $x=-1$ and $x=-5$ .

Solved Example 6

Larger coefficients

Solve

8x^2-22x-21=0

by factorisation.

Show solution

Here $ac=-168$ and $b=-22$ . Use $-28$ and $6$ .

8x^2-28x+6x-21=0

4x(2x-7)+3(2x-7)=0

(2x-7)(4x+3)=0

Roots: $x=\dfrac{7}{2}$ and $x=-\dfrac{3}{4}$ .

Solved Example 7

Consecutive integers problem

The product of two consecutive positive integers is 240. Find the integers.

Show solution

Let the integers be $x$ and $x+1$ .

Then

x(x+1)=240 \Rightarrow x^2+x-240=0

(x+16)(x-15)=0

So $x=-16$ or $x=15$ . Reject the negative value.

The integers are 15 and 16.

Practice Set

Solve by factorisation:

x^2-3x-10=0

2x^2+x-6=0

100x^2-20x+1=0

, and

x^2+x-156=0

Completing the Square

This method converts the quadratic into a perfect square.

How to Think About It

Add just enough to make the left-hand side a perfect square, then take square roots.

Completing the square is especially useful when factorisation is not obvious.

It is also the method from which the quadratic formula is derived.

x^2+bx=\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2

Solved Example 8

Completing the square with rational roots

Solve

9x^2-15x+6=0

by completing the square.

Show solution

Divide by 9:

x^2-\frac{5}{3}x+\frac{2}{3}=0

Move the constant:

x^2-\frac{5}{3}x=-\frac{2}{3}

Add $\left(\dfrac{5}{6}\right)^2$ to both sides:

\left(x-\frac{5}{6}\right)^2=\frac{1}{36}

x-\frac{5}{6}=\pm\frac{1}{6}

Roots: $x=1$ and $x=\dfrac{2}{3}$ .

Solved Example 9

Completing the square with irrational roots

Solve

5x^2-6x-2=0

by completing the square.

Show solution

Divide by 5:

x^2-\frac{6}{5}x=\frac{2}{5}

Add $\left(\dfrac{3}{5}\right)^2$ to both sides:

\left(x-\frac{3}{5}\right)^2=\frac{19}{25}

x-\frac{3}{5}=\pm\frac{\sqrt{19}}{5}

Hence

x=\frac{3\pm\sqrt{19}}{5}

Practice Set

Solve by completing the square:

x^2+4x-1=0

and

2x^2-7x+3=0

Quadratic Formula and Shreedharacharya's Rule

The formula method works for every quadratic equation.

Ancient Indian Mathematics

Shreedharacharya's formula is one of the most celebrated rules in Indian school algebra. It is the all-weather method when factorisation is awkward.

Board-Exam Reminder

Always write down $a$ , $b$ , and $c$ before substituting into the formula. Most mistakes happen because students lose the sign of $b$ or $c$ .

For the equation

ax^2+bx+c=0

, the roots are given directly by the quadratic formula.

In Indian classrooms this formula is often referred to as Shreedharacharya's Rule.

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

Solved Example 10

Direct formula use

Solve

9x^2+7x-2=0

using the quadratic formula.

Show solution

Here $a=9$ , $b=7$ , and $c=-2$ .

D=49-4(9)(-2)=121

x=\frac{-7\pm\sqrt{121}}{18}=\frac{-7\pm11}{18}

So the roots are $x=\dfrac{2}{9}$ and $x=-1$ .

Solved Example 11

Train speed problem

A train covers 300 km at a uniform speed. If its speed were 5 km/h more, the journey would take 2 hours less. Find the original speed.

Show solution

Let speed be $x$ km/h.

\frac{300}{x}-\frac{300}{x+5}=2

300\cdot\frac{5}{x(x+5)}=2

x^2+5x-750=0

x=\frac{-5\pm\sqrt{3025}}{2}=\frac{-5\pm55}{2}

Valid root: $x=25$ km/h.

Solved Example 12

Parameter-based formula problem

Solve

abx^2+(b^2-ac)x-bc=0

using the quadratic formula.

Show solution

Here $A=ab$ , $B=b^2-ac$ , and $C=-bc$ .

D=(b^2-ac)^2+4ab^2c=(b^2+ac)^2

x=\frac{-(b^2-ac)\pm(b^2+ac)}{2ab}

This gives the roots

x=\frac{c}{b}\quad\text{and}\quad x=-\frac{b}{a}

Practice Set

Use the formula to solve:

x^2-3x-10=0

2x^2+x-4=0

, and

5x^2-6x+1=0

Discriminant and Nature of Roots

The discriminant predicts the nature of roots before you fully solve the equation.

Forecast Analogy

The discriminant is like a weather forecast. Even before you solve the equation fully, it tells you what type of result is coming.

The expression

D=b^2-4ac

is called the discriminant.

The sign of the discriminant tells whether the roots are distinct, equal, or not real.

D=b^2-4ac

D>0 \Rightarrow \text{two distinct real roots}

D=0 \Rightarrow \text{two equal real roots}

D<0 \Rightarrow \text{no real roots}

Solved Example 13

State the nature of roots

Find the nature of the roots of (i)

2x^2-x-1=0

, (ii)

x^2+x+1=0

, and (iii)

x^2-4x+4=0

Show solution

For (i),

D=(-1)^2-4(2)(-1)=9>0

so the roots are real and distinct.

For (ii),

D=1-4=-3<0

so there are no real roots.

For (iii),

D=16-16=0

so the roots are real and equal.

Solved Example 14

Find k for equal roots

For what value of

k

does

9x^2-24x+k=0

have equal roots?

Show solution

For equal roots,

D=0

(-24)^2-4(9)k=0

576-36k=0 \Rightarrow k=16

Solved Example 15

Find k for real roots

Find the values of

k

for which

kx^2-6x-2=0

has real roots.

Show solution

For real roots,

D\ge0

36+8k\ge0 \Rightarrow k\ge-\frac{9}{2}

Also $k\ne0$ so that the equation remains quadratic.

Practice Set

1. Find

k

2x^2+kx+3=0

has equal roots.
2. State the nature of roots of

x^2-5x+6=0

x^2+2x+5=0

, and

x^2-2x+1=0

Word Problems and Applications

Choose one variable carefully, translate the story, and then solve the quadratic.

Common Patterns

Consecutive numbers: $x$ and $x+1$
Rectangle dimensions: breadth $=x$ , length $=x+k$
Age problems: present age $=x$ , future age $=x+n$
Speed problems: time $=\dfrac{\text{distance}}{\text{speed}}$
Digit problems: two-digit number $=10x+y$

Exam Tip

Always reject impossible roots after solving. Negative speed, negative age, and invalid digits cannot be final answers.

Age, number, geometry, train, boat, rectangle, digit, and budget problems often reduce naturally to quadratics.

The hard part is usually the model, not the algebra. Once the equation is formed correctly, the rest becomes methodical.

Solved Example 16

Age problem

Rohan's mother is 26 years older than him. The product of their ages 3 years from now will be 360. Find Rohan's present age.

Show solution

Let Rohan's age be $x$ years. Then his mother's age is $x+26$ .

After 3 years, their ages are $x+3$ and $x+29$ .

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

x^2+32x-273=0

x=\frac{-32\pm46}{2}

So $x=7$ or $x=-39$ . Reject the negative value. Rohan is 7 years old.

Solved Example 17

Right triangle dimensions

The area of a right triangle is

600\text{ cm}^2

. Its base is 10 cm more than its altitude. Find both dimensions.

Show solution

Let altitude be $x$ cm, so base is $x+10$ cm.

\frac{1}{2}x(x+10)=600

x^2+10x-1200=0

(x-30)(x+40)=0

Reject the negative value. Altitude = 30 cm and base = 40 cm.

Solved Example 18

Boat and stream problem

A boat in still water has speed 15 km/h. It goes 30 km upstream and returns downstream in 4.5 hours. Find the speed of the stream.

Show solution

Let stream speed be $x$ km/h. Then upstream speed is $15-x$ and downstream speed is $15+x$ .

\frac{30}{15-x}+\frac{30}{15+x}=\frac{9}{2}

\frac{900}{225-x^2}=\frac{9}{2}

1800=2025-9x^2 \Rightarrow x^2=25

So $x=5$ . The stream speed is 5 km/h.

Solved Example 19

Marbles problem

John and Jivanti together have 45 marbles. Both lose 5 marbles each, and the product of the marbles they now have is 128. How many marbles did they have originally?

Show solution

Let John have $x$ marbles. Then Jivanti has $45-x$ marbles.

After losing 5 each, they have $x-5$ and $40-x$ marbles.

(x-5)(40-x)=128

x^2-45x+328=0

(x-36)(x-9)=0

So the original counts were 36 and 9 marbles, in either order.

Board-Style Practice

1. Find two consecutive positive integers whose squares add up to 365.
2. The sum of two numbers is 15 and the sum of their reciprocals is

\dfrac{3}{10}

. Find the numbers.
3. A train travels 360 km at uniform speed. If the speed were 5 km/h more, the journey would take 1 hour less. Find the speed.
4. A rectangular park has perimeter 82 m and area 400 m

^2

. Find its breadth.
5. A two-digit number has product of digits 18. If 63 is subtracted from it, the digits interchange. Find the number.
6. The sum of first

n

even natural numbers is 420. Find

n

Quick Summary

Concept	Key Idea
Standard form	$ax^2+bx+c=0$ where $a\ne0$ .
Root	A value of $x$ that makes the expression equal to 0.
Factorisation	Best when the expression splits neatly into linear factors.
Completing the square	Turns the quadratic into a perfect square.
Quadratic formula	$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ .
Discriminant	$D=b^2-4ac$ predicts the nature of roots.
Nature of roots	$D>0$ distinct, $D=0$ equal, $D<0$ no real roots.
Sum and product of roots	$\alpha+\beta=-\dfrac{b}{a}$ and $\alpha\beta=\dfrac{c}{a}$ .

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