# Quadratic Equations Class 10 Notes Maths Chapter 4

## What is a Quadratic  Equation?

A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0 is called the standard form of a quadratic equation.

Example 1: Check whether the following are quadratic equations

i) (x + 1)2 = 2(x − 3)
⇒ (x + 1)2 = x2 + 2x + 1
∵ (a + b)2 = a2 + 2ab + b2

⇒ x2 + 2x + 1 = 2(x − 3)
⇒ x2 + 2x + 1 = 2x − 6

⇒ x2 + 2x + 1 − 2x + 6 = 0
⇒ x2 + 2x − 2x + 6 + 1 = 0
x2 + 7 = 0
The above equation is a quadratic equation, where the coefficient of x is zero, i.e. b = 0

ii) x(x + 1)(x + 8) = (x + 2)(x − 2)
LHS
⇒ x(x + 1)(x + 8)
⇒ x(x2 + 8x + x + 8)

⇒ x(x2 + 9x + 8)
⇒ x3 + 9x2 + 8x

RHS
(x + 2)(x − 2)
⇒ x2 − 4
∵ (a + b)(a − b) = a2 − b2
Now, x3 + 9x2 + 8x = x2 − 4
⇒ x3 + 9x2 − x2 + 8x + 4 = 0
x3 + 8x2 + 8x + 4 = 0
It is not a quadratic equation as it is an equation of degree 3.

iii) (x − 2)+ 1 = 2x − 3
LHS
(x − 2)2 + 1 = x2 − 2x + 4 + 1
∵ (a − b)2 = a2 − 2ab + b2
= x2 − 2x + 5
RHS
2x − 3
⇒ x2 − 2x + 5 = 2x − 3
⇒ x2 − 2x − 2x + 5 + 3 = 0
⇒ x2 − 4x + 8 = 0
The above equation is quadratic as it is of the form,
ax2 + bx + c = 0

Example 2: The product of two consecutive positive integers is 420. Form the equation satisfying this scenario.

Let the two consecutive positive integers be x and x + 1 Product of the two consecutive integers= x(x + 1) = 420

⇒ x2 + x = 420

⇒ x+ x − 420 = 0

x2 + x − 420 = 0, is the required quadratic equation and the two integers satisfy this quadratic equation.

Question for Chapter Notes: Quadratic Equations
Try yourself:Which of the following is the correct quadratic equation of the expression (x + 3)(x - 2)?

### Solution of Quadratic Equation by Factorisation

A real number α is called a root of the quadratic equation x+bx + c = 0 , a ≠ 0 if aα2 + bα + c = 0
We say that x = α is a solution of the quadratic equation.
Example: x2 − 2x − 3 = 0
If we put x = −1 in the LHS of the above equation we get,
(−1)2 − 2(−1) − 3
1 + 2 − 3 = 0
Thus x = −1 is a solution of the equation x2 − 2x − 3 = 0.
To find the roots of the quadratic equations we follow these steps.

Example 3: Find the roots of the equation x2 − 3x − 10 = 0

The given equation is x2 − 3x − 10 = 0.
Here a = 1, b = −3 and c = −10
1) Find the product of a and c .
Here, the product of a and c= -10 → (ac) is negative

2) Write the factors of this product (ac) such that the sum of the two factors is equal to b.
∴ ac = m × n and m + n = b
Factors of 10 = 2×5
Let m = −5 and n = 2 → (ac=-10)
We write the given equation as,

x2 − 5x + 2x − 10 = 0
x(x − 5) + 2(x − 5) = 0
x − 5 x + 2 = 0
Equate each factor to zero to get the roots of the equation.
x − 5 = 0 and x + 2 = 0
x = 5, −2
Therefore, 5 and -2 are the roots of the equation x2 − 3x − 10 = 0

Example 4: Solve the following quadratic equation by factorisation method.

i) 4√3x2  + 5x − 2√3  = 0
The given equation is 4√3x2  + 5x − 2√3  = 0

Here, a = 4√3 , b = 5 and c = −2√3

The product of a and c
= 4√3  × (−2√3 )
= −8 × 3
= −24
Factors of 24 = 3×8 and 8 + (−3) = 5
The factors of the equation are 8, − 3
So, the given equation can be written as,

4√3x2  + (8 − 3)x − 2√3  = 0
⇒ 4 √3x2 + 8x − 3x − 2√3  = 0

⇒ 4x( √3x + 2) −√3 (√3 x + 2) = 0
⇒ (4x − √3 )(√3x + 2) = 0
Equating each factor to zero we get,

(4x −√3 )=0 and (√3x + 2) = 0

The roots of the equation

The given equation is
Multiplying the above equation by xwe get,

Here, a = 2, b = −5 and c = 2
The product of a and c = 2 × 2 = 4
The factors of 4 = 4 × 1 and 4 + 1 = 5
2x2 − (4 + 1)x + 2 = 0
⇒ 2x2 − 4x − 1x + 2 = 0
2x(x − 2) − (x − 2) = 0
(2x − 1)(x − 2) = 0
Equating each factor to zero we get,
(2x − 1) = 0 and (x − 2) = 0
x =  1/2 and x = 2

The roots of equation 2x− 5x + 2 = 0 are  1/2  and 2

Example 5: The altitude of a right-angled triangle is 7 cm less than its base. If the hypotenuse is 13 cm long, then find the other two sides.

Let the length of the base be x cm, then altitude = x − 7 cm

Hypotenuse = 13 cm

We know, H2 = P2 + B2
132 = (x − 7)2 + x2
⇒ 169 = x2 − 14x + 49 + x2
⇒ x2 − 14x + 49 + x2 = 169
⇒ 2x2 − 14x + 49 − 169 = 0
⇒ 2x2 − 14x − 120 = 0
Dividing the above equation by 2 we get,

x2 − 7x − 60 = 0
Here, a = 1, b = −7 and c = −60
The product of a and c = 1 × (−60) = −60
The factors of 60 = 5 × 12 and −12 + 5 = 7
The given equation can be written as,

x2 − 12x + 5x − 60 = 0
x(x − 12) + 5(x − 12) = 0
⇒ (x + 5)(x − 12) = 0
Equating each factor to zero we get,
(x + 5) = 0 and (x − 12) = 0
⇒ x = −5 and x = 12
The length of the base cannot be negative.
Therefore, Base = 12 cm
Altitude = x − 7 cm = 12 − 7 = 5 cm, Hypotenuse = 13 cm

Question for Chapter Notes: Quadratic Equations
Try yourself:Which of the following is the solution(s) of the quadratic equation x^2 + 5x + 6 = 0 by factorisation method?

### Nature of Roots

The roots of the quadratic equation ax2 + bx + c = 0 are given by

Where D = √b2 − 4ac is called the discriminant.

This formula is known as the Quadratic Formula.

The nature of the roots depends upon the value of Discriminant, D.

Example 6: Find the roots of the equation,

The given equation is

Squaring both sides of the equation we get,

Here, a = 4, b = −37 and c = 40

Substituting the value of a, b and c in the quadratic formula

Taking +ve sign first,

Taking -ve we get,

The roots of the given equation are 8 and 5/4.

Example 7: Find the numerical difference of the roots of the equation x2 − 7x − 30 = 0

The given quadratic equation is x− 7x − 30 = 0

Here a = 1, b = −7 and c = −30

Substituting the value of a, b and c in the quadratic formula

Taking +ve sign first,

Taking -ve we get,

The two roots are 10 and -3

The difference of the roots= 10 − (−3) = 10 + 3 = 13

Example 8: Find the discriminant of the quadratic equation x2 −4x − 5 = 0

The given quadratic equation is x2 − 4x − 5 = 0.

On comparing with ax2 + bx + c = 0 we get,

a = 1,b = −4, and c = −5

Example 9: Find the value of p, so that the quadratic equation px(x − 2) + 9 = 0 has equal roots.

The given quadratic equation is px(x − 2) + 9 = 0
px2 − 2px + 9 = 0

Now comparing with ax2 + bx + c = 0 we get,

a = p, b = −2p and c = 9

The given quadratic equation will have equal roots if D = 0

p = 0 and p − 9 = 0 ⇒ p = 9
p = 0 and p = 9
The value of p cannot be zero as the coefficient of x, (−2p) will become zero.
Therefore, we take the value of p = 9.

Example 10: If x = −1 is a root of the quadratic equations 2x2 +px + 5 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the value of k.

The given quadratic equation is 2x2 + px + 5 = 0. If x = −1 is the root of the equation then,
2(−1)2 + p(−1) + 5 = 0
2 − p + 5 = 0
⇒ −p = −7
p = 7
Putting the value of p in the equation p(x2 + x) + k = 0,
7(x2 + x) + k = 0
⇒ 7x2 + 7x + k = 0
Now comparing with ax2 + bx + c = 0 we get,
a = 7, b = 7 and c = k

The given quadratic equation will have equal roots if D = 0

Therefore, the value of k is 7/4.

Question for Chapter Notes: Quadratic Equations
Try yourself:What is the nature of the roots of the quadratic equation 2x² + 5x + 3 = 0?

### Solution of the Quadratic Equations by Completing the  Square (Deleted from NCERT Textbook)

If we have to find the solution of a quadratic equation by completing the square, we follow the steps given below.

The complete square is,

Example 11: Find the roots of the following quadratic equations by the method of completing the square:

i) 2x2 − 7x + 3 = 0

The given quadratic equation is 2x2 − 7x + 3 = 0

The coefficient of x2 is not 1, so we divide the whole equation by 2.

Now move 3/2 to RHS

Adding (7/4)to both sides we get,

Taking square root of both sides we get,

The roots of the equation are 3 and 1/2

ii) 4x+ 4√3x + 3 = 0

Dividing the whole equation by 4, so that the coefficient of x2 is 1.

Shifting 3/4 to RHS

Adding  to both sides we get,

Taking the square root of both sides

The roots of the given equation are

Example 12: Solve the quadratic equation

x2 − (√5 + 1)x + √5 = 0 by completing the square method.
The given quadratic equation is, x2 − (√5 + 1)x = −√5 = 0

Shifting √5 to RHS we get,

x2 − (√5 + 1)x = −√5
Adding  to both sides we get,

Taking +ve sign first

Taking –ve sign

The roots of the given equation are 5 and 1

The document Quadratic Equations Class 10 Notes Maths Chapter 4 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

## Mathematics (Maths) Class 10

120 videos|463 docs|105 tests

## FAQs on Quadratic Equations Class 10 Notes Maths Chapter 4

 1. What is a quadratic equation?
Ans. A quadratic equation is a polynomial equation of the second degree, which can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants and a ≠ 0. It represents a curve called a parabola.
 2. How can we solve a quadratic equation by factorisation?
Ans. To solve a quadratic equation by factorisation, we need to factorise the quadratic expression and equate each factor to zero to find the values of x. The solutions obtained by this method are called the roots of the quadratic equation.
 3. What is the nature of roots in a quadratic equation?
Ans. The nature of roots in a quadratic equation depends on the discriminant (D) of the equation. If D > 0, the equation has two distinct real roots. If D = 0, the equation has two equal real roots. If D < 0, the equation has two complex roots.
 4. How can we solve a quadratic equation by completing the square?
Ans. To solve a quadratic equation by completing the square, we follow these steps: 1. Write the equation in the form ax^2 + bx + c = 0. 2. Divide the entire equation by the coefficient of x^2 to make the coefficient of x^2 equal to 1. 3. Complete the square by adding and subtracting (b/2a)^2 to the equation. 4. Rewrite the equation in the form (x + p)^2 = q, where p and q are constants. 5. Take the square root of both sides and solve for x.
 5. Why was the method of solving quadratic equations by completing the square deleted from the NCERT textbook?
Ans. The method of solving quadratic equations by completing the square was deleted from the NCERT textbook to simplify the curriculum and focus on more important concepts. Additionally, this method involves more steps and calculations compared to other methods like factorisation and using the quadratic formula.

## Mathematics (Maths) Class 10

120 videos|463 docs|105 tests

### Up next

 Explore Courses for Class 10 exam

### Top Courses for Class 10

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Track your progress, build streaks, highlight & save important lessons and more!
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;