Class 10 Chapter 4 (Quadratic Equations)

Notes on Quadratic Equations

Quadratic Polynomial

A polynomial, whose degree is 2, is called a quadratic polynomial. It is in the form of

p(x) = ax² + bx + c, where a ≠ 0

Quadratic Equation

When we equate the quadratic polynomial to zero then it is called a Quadratic Equation i.e. if

p(x) = 0, then it is known as Quadratic Equation. 

Standard form of Quadratic Equation

where a, b, c are the real numbers and a≠0

Types of Quadratic Equations

1. Complete Quadratic Equation  ax² + bx + c = 0, where a ≠ 0, b ≠ 0, c ≠ 0

2. Pure Quadratic Equation   ax² = 0, where a ≠ 0, b = 0, c = 0

Roots of a Quadratic Equation

Let x = α where 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 2, therefore Quadratic Equations can have two roots. So the zeros of quadratic polynomial p(x) = ax² + bx + c is same as the roots of the Quadratic Equation ax²+ bx + c = 0.

Methods to solve the Quadratic Equations

There are three methods to solve the Quadratic Equations-

1. Factorization Method

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

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

Step 2: Split the middle term bx as mx + nx so that the sum of m and n is equal to b and the product of m and n is equal to ac.

Step 3: By factorization we get the two linear factors (x + p) and (x + q)

ax² + bx + c = 0 = (x + p) (x + q) = 0

Step 4: Now we have to equate each factor to zero to find the value of x.

\[ \Rightarrow {x^2} - 2x - 15 = 0\]

\[(x + 3)(x - 5) = 0\]

\[x + 3 = 0\,\,or\,\,x - 5 = 0\]

\[x =  - 3\,\,or\,\,x = 5\]

These values of x are the two roots of the given Quadratic Equation.

2. Completing the square method

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

Step1: Given Quadratic Equation in the standard form ax² + bx + c = 0.

Step 2: Divide both sides by a

\[{x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0\]

Step 3: Transfer the constant on RHS then add square of the half of the coefficient of x i.e. \({\left( {\dfrac{b}{{2a}}} \right)^2}\) on both sides

\[{x^2} + \dfrac{b}{a}x =  - \dfrac{c}{a}\]

\[{x^2} + \dfrac{b}{a}x + {\left( {\dfrac{b}{{2a}}} \right)^2} =  - \dfrac{c}{a} + {\left( {\dfrac{b}{{2a}}} \right)^2}\]

Step 4: Now write LHS as perfect square and simplify the RHS.

\[{\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{{b^2} - 4ac}}{{4{a^2}}}\]

Step 5: Take the square root on both the sides.

\[x + \dfrac{b}{{2a}} =  \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}} \]

Step 6: Now shift all the constant terms to the RHS and we can calculate the value of x as there is no variable at the RHS.

\[x =  \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}}  - \dfrac{b}{{2a}}\]

3. Quadratic formula method

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

\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

where a, b and c are the real numbers and b² – 4ac is called discriminant.

To find the roots of the equation, put the value of a, b and c in the quadratic formula.

Nature of Roots

From the quadratic formula, we can see that the two roots of the Quadratic Equation are -

\[x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}\,\,and\,\,x = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}\]

Or \(x = \dfrac{{ - b + \sqrt D }}{{2a}}\) ; Where D = b² – 4ac

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

∆ = Discriminant