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MODULE - 1 Quadratic Equations
Algebra
Notes 6
QUADRATIC EQUATIONS
In this lesson, you will study about quadratic equations. You will learn to identify quadratic
equations from a collection of given equations and write them in standard form. You will
also learn to solve quadratic equations and translate and solve word problems using quadratic
equations.
OBJECTIVES
After studying this lesson, you will be able to
• identify a quadratic equation from a given collection of equations;
• write quadratic equations in standard form;
• solve quadratic equations by (i) factorization and (ii) using the quadratic formula;
• solve word problems using quadratic equations.
EXPECTED BACKGROUND KNOWLEDGE
• Polynomials
• Zeroes of a polynomial
• Linear equations and their solutions
• Factorisation of a polynomial
6.1 QUADRATIC EQUATIONS
You are already familiar with a polynomial of degree two. A polynomial of degree two is
called a quadratic polynomial. When a quadratic polynomial is equated to zero, it is called
a quadratic equation. In this lesson, you will learn about quadratic equations in one
variable only. Let us consider some examples to identify a quadratic equation from a
collection of equations
170 Mathematics Secondary Course
Quadratic Equations MODULE - 1
Algebra
Example 6.1: Which of the following equations are quadratic equations?
2 2
(i) 3x = 5 (ii) x + 2x + 3 = 0
3 2
(iii) x + 1 = 3x (iv) (x + 1) (x + 3) = 2x + 1 Notes
(v) x + 1 = 5 (v) x2 + x +1=0
x 2
Solution:
2 2 2
(i) It is a quadratic equation since 3x = 5 can be written as 3x – 5 = 0 and 3x – 5 is a
quadratic polynomial.
(ii) x2 2
+ 2x + 3 = 0 is a quadratic equation as x + 2x + 3, is a polynomial of degree 2.
3 2 3 2
(iii) x + 1 = 3x can be written as x – 3x + 1 = 0. LHS is not a quadratic polynomial
since highest exponent of x is 3. So, the equation is not a quadratic equation.
(iv) (x + 1) (x + 3) = 2x + 1 is a quadratic equation, since (x + 1) (x + 3) = 2x + 1 can be
written as
2
x + 4x + 3 = 2x + 1
2
or x + 2x + 2 = 0
Now, LHS is a polynomial of degree 2, hence (x + 1) (x + 3) = 2x + 1 is a quadratic
equation.
(v) x+ 1 = 5 is not a quadratic equation.
x 2
However, it can be reduced to quadratic equation as shown below:
x+1 = 5
x 2
x2 +1 5
or x = 2,x ≠ 0
2
or 2(x + 1) = 5x , x ≠ 0
2
or 2x – 5x + 2 = 0, x ≠ 0
(vi) x2 + x +1=0 is not a quadratic equation as x2 + x +1 is not a quadratic
polynomial (Why?)
Mathematics Secondary Course 171
MODULE - 1 Quadratic Equations
Algebra
CHECK YOUR PROGRESS 6.1
Notes 1. Which of the following equations are quadratic equations?
2 3 2
(i) 3x + 5 = x + x (ii) 3x + 5x + 2 = 0
x2 +1 5
(iii) (5y + 1) (3y – 1) = y + 1 (iv) x +1 = 2
2
(v) 3x + 2x = 5x – 4
6.2 STANDARD FORM OF A QUADRATIC EQUATION
2
A quadratic equation of the form ax + bx + c = 0, a > 0 where a, b, c, are constants and
x is a variable is called a quadratic equation in the standard form. Every quadratic equation
can always be written in the standard form.
Example 6.2: Which of the following quadratic equations are in standard form? Those
which are not in standard form, express them in standard form.
2 2
(i) 2 + 3x + 5x = 0 (ii) 3x – 5x + 2 = 0
2
(iii) 7y – 5y = 2y + 3 (iv) (z + 1) (z + 2) = 3z + 1
2
Solution: (i) It is not in the standard form. Its standard form is 5x + 3x + 2 = 0
(ii) It is in standard form
(iii) It is not in the standard form. It can be written as
2
7y – 5y = 2y + 3
2
or 7y – 5y – 2y – 3 = 0
2
or 7y – 7y – 3 = 0
which is now in the standard form.
(iv) It is not standard form. It can be rewritten as
(z + 1) (z + 2) = 3z + 1
or z2
+ 3z + 2 = 3z + 1
2
or z + 3z – 3z + 2 – 1 = 0
or z2
+ 1 = 0
2
or z + 0z + 1 = 0
which is now in the standard form.
172 Mathematics Secondary Course
Quadratic Equations MODULE - 1
Algebra
CHECK YOUR PROGRESS 6.2
1. Which of the following quadratic equations are in standard form? Those, which are Notes
not in standard form, rewrite them in standard form:
2 2
(i) 3y – 2 = y + 1 (ii) 5– 3x – 2x = 0
(iii) (3t – 1) (3t + 1) = 0 (iv) 5 – x = 3x2
6.3 SOLUTION OF A QUADRATIC EQUATION
You have learnt about the zeroes of a polynomial. A zero of a polynomial is that real
number, which when substituted for the variable makes the value of the polynomial zero. In
case of a quadratic equation, the value of the variable for which LHS and RHS of the
equation become equal is called a root or solution of the quadratic equation. You have also
learnt that if α is a zero of a polynomial p(x), then (x –α) is a factor fo p(x) and conversely,
if (x –α) is a factor of a polynomial, then α is a zero of the polynomial. You will use these
results in finding the solution of a quadratic equation. There are two algebraic methods for
finding the solution of a quadratic equation. These are (i) Factor Method and (ii) Using the
Quadratic Formula.
Factor Method
Let us now learn to find the solutions of a quadratic equation by factorizing it into linear
factors. The method is illustrated through examples.
Example 6.3: Solve the equation (x – 4)(x + 3) = 0
Solution: Since, (x – 4)(x + 3) = 0, therefore,
either x – 4 = 0, or x + 3 = 0
or x = 4 or x = – 3
Therefore, x = 4 and x = – 3 are solutions of the equation.
2
Example 6.4: Solve the equation 6x + 7x – 3 = 0 by factorisation.
2
Solution: Given 6x + 7x – 3 = 0
By breaking middle term, we get
2
6x + 9x – 2x – 3 = 0 [since, 6 × (–3) = – 18 and – 18 = 9 × (–2) and 9 – 2 = 7]
or 3x(2x + 3) – 1(2x + 3) = 0
or (2x + 3) (3x – 1) = 0
This gives 2x + 3 = 0 or 3x – 1 = 0
or x = 3 or x = 1
2 3
Mathematics Secondary Course 173
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