Quadratic Equations Class 10th Notes - Free NCERT Class 10 Maths Chapter 4 Notes - Download PDF

Imagine after building your dream house, you need to fence that house, and for that, you need the length and breadth of that house to find the biggest possible area. Quadratic equations can solve this problem. It is an integral part of Mathematics that deals with algebraic equations of degree 2. These mathematical equations function beyond theory because they serve various purposes in real-life scenarios. People use quadratic equations in all parts of life to determine areas and moving object speeds, and make financial predictions while resolving problems throughout physics and engineering. This section creates the fundamental base required to develop algebraic reasoning abilities together with problem-solving capability. The ability to identify and solve quadratic equations will help to succeed in the examinations and ensure success in their subsequent educational and competitive examination programs.

This Story also Contains

1. Quadratic Polynomial
2. Quadratic Equation
3. Roots of a Quadratic Equation
4. Methods to Solve Quadratic Equations
5. Nature of Roots
6. Solving using the quadratic formula when D > 0:
7. Class 10 Chapter Wise Notes
8. NCERT Exemplar Solutions for Class 10
9. NCERT Solutions for Class 10
10. NCERT Books and Syllabus

The NCERT Class 10 Maths Chapter 4 notes cover a brief outline of the chapter on quadratic equations and can be used for revision. The main topics covered in the quadratic equations Class 10 notes give you the properties and roots of quadratic equations. Class 10 Maths chapter 4 notes also cover the basic equations in the chapter. Quadratic equations Class 10 notes PDF download contains all of these topics. Apart from this, the use of NCERT class 10th maths notes provides you with the ability to examine mathematical concepts rigorously. Also, after studying the NCERT Notes, you will have access to short summaries that help you to remember information fast.

Quadratic Polynomial

A polynomial with degree 2 is a quadratic polynomial. It is in the form of: f(x) = ax² + bx + c, where a ≠ 0.

Quadratic Equation

A quadratic polynomial, when equated to a constant (like 0), then the equation becomes a Quadratic Equation, which means f(x) = 0.

The standard form of a Quadratic Equation is: f(x) = ax² + bx + c = 0, where a, b and c are the real numbers and a ≠ 0.
Here, a is called as quadratic coefficient, as it is the coefficient of x² and b is called as linear coefficient, as it is the coefficient of x, and c is the constant term.

Roots of a Quadratic Equation

A quadratic equation's roots represent x values that satisfy the given equation.
Let x = α, and α is a real number. If α satisfies the quadratic equation ax²+ bx + c = 0 such that aα² + bα + c = 0, then α is the root of the Quadratic Equation.

• As quadratic polynomials have degree two, quadratic equations can have two roots. Thus, the zeros of a quadratic polynomial f(x) = ax² + bx + c are the same as the roots of the quadratic equation ax² + bx + c = 0.
• Roots can be of three types, in case of a quadratic equation: two distinct roots, two equal roots or real roots may not exist.

Methods to Solve Quadratic Equations

There are three methods to solve Quadratic Equations.

1. Factorization Method

In this method, factorize the equation into two linear factors and equate each factor to zero to find the roots of the equation.

Step 1: Quadratic Equation in the form of ax² + bx + c = 0.

Step 2: Find the two numbers (let's say p and q), whose sum is equal to 'b' and product is equal to 'a × c'.

Step 3: By factorization, write ax² + bx + c = 0 as (x + p) (x + q) = 0

For example-

x²- 2x - 15=0

⇒ (x+3)(x-5)=0

So, x + 3 = 0 or, x - 5 = 0

$\therefore$ x = - 3 or x = 5

The above values of x are the two roots of the given quadratic equation.

2. Completing the Square Method

In this method, convert the equation to square form (x + a)² - b² = 0 or (x + a)² = b² to find the roots.

Example: Solve $ x^2 + 6x + 5 = 0 $

Solution:

• Step 1: Make sure the coefficient of $x²$ is 1
  In this case, it's already 1.
  If not, divide the equation with the coefficient of $x²$.

• Step 2: Move the constant term to the right-hand side:
  $x^2 + 6x = -5$

• Step 3: Add the square of half the coefficient of $x$ to both sides:
  Half of 6 is 3, and $3^2 = 9$
  So, $x^2 + 6x + 9 = -5 + 9$

• Step 4: Write the left-hand side as a perfect square, to make the form (x + a)² = b²
  $(x + 3)^2 = 4$

• Step 5: Take the square root of both sides:
  $x + 3 = \pm \sqrt{4}$

• Step 6: Solve for $x$:
  $x + 3 = \pm 2$
  $x = -3 \pm 2$

Therefore, $x$ = -1 or -5

3. Quadratic formula method

In this method, find the roots by using the quadratic formula. The quadratic formula is

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

where a, b and c are the real numbers and b² – 4ac is known as the discriminant. In this method, we directly get the roots once we substitute the values in the formula.

Example: Solve $ 2x^2 - 4x - 6 = 0$

Solution: Here, a = 2, b = -4, c = -6
Apply the quadratic formula: $x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

$ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2(2)}$

$ x = \frac{4 \pm \sqrt{16 + 48}}{4}$

$ x = \frac{4 \pm \sqrt{64}}{4}$

$x = \frac{4 \pm 8}{4}$

$ x = \frac{4 + 8}{4} = 3 \quad \text{or} \quad x = \frac{4 - 8}{4} = -1$

$x = 3 \quad \text{or} \quad x = -1$

Nature of Roots

The nature of the roots of the equation depends upon the value of D, which is called the discriminant.

Solving using the quadratic formula when D > 0:

When the quadratic equation has D > 0, the roots of the equation are distinct and real.

Example: Solve $ x^2 - 5x + 6 = 0 $

Solution: Here, a = 1, b = -5, c = 6

D = $(-5)^2 - 4(1)(6)$ = 25 - 24 = 1

It means D > 0.

Now, apply the quadratic formula:

x = $\frac{-(-5) \pm \sqrt{1}}{2(1)} = \frac{5 \pm 1}{2}$

x = $3 \quad \text{or} \quad x$ = 2

Therefore, the roots are distinct and real.

Solving using the quadratic formula when D = 0:

When the quadratic equation has D = 0, the roots of the equation are equal.

Example: Solve $ x^2 - 4x + 4 = 0 $

Solution: Here, a = 1, b = -4, c = 4

D = $(-4)^2 - 4(1)(4) $= 16 - 16 = 0

It means D = 0.

Now, apply the quadratic formula:

x = $\frac{-(-4)}{2(1)} = \frac{4}{2}$ = 2

Therefore, x = 2

Thus, the roots are equal, as we get only one root.

Solving using the quadratic formula when D < 0:

When the quadratic equation has D < 0, the roots of the equation are not real (complex).

Example: Solve $ x^2 + 2x + 5 = 0 $

Solution: Here, a = 1, b = 2, c = 5

D = $(2)^2 - 4(1)(5)$ = 4 - 20 = -16

It means D < 0.

Now, apply the quadratic formula:

x = $\frac{-2 \pm \sqrt{-16}}{2(1)} = \frac{-2 \pm 4i}{2}$

x = $-1 \pm 2i$

Thus, the roots are not real and are complex numbers.

Class 10 Chapter Wise Notes

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NCERT Exemplar Solutions for Class 10

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NCERT Solutions for Class 10

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NCERT Books and Syllabus

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• NCERT Books for Class 10
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