Problems on Quadratic Equations

We will discuss here about some of the problems on quadratic equations.

1. Solve: x^2 = 36

x^2 = 36

or, x^2 - 36=0

or, (x + 6)(x - 6) = 0

So, one of x + 6 and x - 6 must be zero

From x + 6 = 0, we get x = -6

From x - 6 = 0, we get x = 6

Thus, the required solutions are x = ± 6

Keeping the expression involving the unknown quantity and the constant term on left and right side respectively and finding square root from both sides, we can solve the equation also.

As in the equation x^2 = 36, finding square root from both sides, we get x = ± 6.


2.  Solve 2x^2 - 5x + 3 = 0

2x^2 - 5x + 3 = 0

or 2x^2 - 3x – 2x + 3=0

or, x (2x - 3) - 1 (2x - 3)=0

or, (x - 1)(2x - 3) = 0

Therefore, one of (x - 1) and (2x - 3) must be zero.

when, x - 1 = 0, x = 1

and when 2x - 3 = 0, x = 3/2

Thus required solutions are x = 1, 3/2

 

3. Solve: 3x^2 - x = 10

3x^2 - x = 10

or, 3x^2 - x - 10 = 0

or, 3x^2 - 6x + 5x - 10 = 0

or, 3x (x - 2) + 5 (x - 2) =0

or, (x - 2)(3x + 5) = 0

Therefore, one of x - 2 and 3x + 5 must be zero

When x - 2 = 0, x = 2

and when 3x + 5 = 0; 3x = -5 or; x = -5/3

Therefore, required solutions are x= -5/3, 2

 

4. Solve: (x - 7)(x - 9) = 195

(x - 7)(x - 9) = 195

or, x^2 - 9x – 7x + 63 – 195 = O

or, x2 - 16x - 132=0

or, x^2 - 22 x + 6x - 132=0

or, x(x - 22) + 6(x - 22) = 0

or, (x - 22)(x + 6) = 0

Therefore, one of x - 22 and x + 6 must be zero.

When x - 22, x = 22

when x + 6 = 0, x = - 6

Required solutions are x= -6, 22



5. Solve: x/3 +3/x =  4 1/4

or, x2 + 9/3x = 17/4

or, 4x2 + 36 = 51x

or, 4x^2 - 51x + 36 = 0

or, 4x^2 - 48x - 3x + 36 = 0

or, 4x(x- 12) -3(x - 12) = 0

or, (x - 12)(4x -3) = 0

Therefore, one of (x - 12) and (4x - 3) must be zero.

When x - 12 = 0, x = 12 when 4x -3 = 0,x = 3/4


6. Solve: x - 3/x + 3 - x + 3/x - 3 + 6 6/7 = 0

Assuming x - 3/x + 3 = a, the given equation can be written as:

a - 1/a + 6 6/7 = 0

or, a2 - 1/a + 48/7 = 0

or, a2 - 1/a = - 48/7

or, 7a^2 - 7 = - 48a

or, 7a^2 + 48a - 7 = 0

or,7a^2 + 49a - a - 7 = 0

or, 7a(a + 7) - 1 (a + 7) = 0

or,(a + 7)(7a - 1) = 0

Therefore, 0ne of (a + 7) and (7a - 1) must be zero.

a + 7 = 0 gives a = -7 and 7a - 1 = 0 gives a = 1/7

From a = -7 we get x -3/x + 3 = -7

or, x – 3 = -7x - 2 1

or, 8x = -18

Therefore, x = -18/8 = - 9/4

Again, from a = 1/7, we get x - 3/x + 3 = 1/ 7

or, 7x - 21 = x + 3

or,6x = 24

Therefore, x = 4

Required solutions are x = -9/4, 4

Quadratic Equation

Introduction to Quadratic Equation

Formation of Quadratic Equation in One Variable

Solving Quadratic Equations

General Properties of Quadratic Equation

Methods of Solving Quadratic Equations

Roots of a Quadratic Equation

Examine the Roots of a Quadratic Equation

Problems on Quadratic Equations

Quadratic Equations by Factoring

Word Problems Using Quadratic Formula

Examples on Quadratic Equations 

Word Problems on Quadratic Equations by Factoring

Worksheet on Formation of Quadratic Equation in One Variable

Worksheet on Quadratic Formula

Worksheet on Nature of the Roots of a Quadratic Equation

Worksheet on Word Problems on Quadratic Equations by Factoring








9th Grade Math

From Problems on Quadratic Equations to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Method of H.C.F. |Highest Common Factor|Factorization &Division Method

    Apr 13, 24 05:12 PM

    HCF by Short Division Method
    We will discuss here about the method of h.c.f. (highest common factor). The highest common factor or HCF of two or more numbers is the greatest number which divides exactly the given numbers. Let us…

    Read More

  2. Factors | Understand the Factors of the Product | Concept of Factors

    Apr 13, 24 03:29 PM

    Factors
    Factors of a number are discussed here so that students can understand the factors of the product. What are factors? (i) If a dividend, when divided by a divisor, is divided completely

    Read More

  3. Methods of Prime Factorization | Division Method | Factor Tree Method

    Apr 13, 24 01:27 PM

    Factor Tree Method
    In prime factorization, we factorise the numbers into prime numbers, called prime factors. There are two methods of prime factorization: 1. Division Method 2. Factor Tree Method

    Read More

  4. Divisibility Rules | Divisibility Test|Divisibility Rules From 2 to 18

    Apr 13, 24 12:41 PM

    Divisibility Rules
    To find out factors of larger numbers quickly, we perform divisibility test. There are certain rules to check divisibility of numbers. Divisibility tests of a given number by any of the number 2, 3, 4…

    Read More

  5. Even and Odd Numbers Between 1 and 100 | Even and Odd Numbers|Examples

    Apr 12, 24 04:22 PM

    even and odd numbers
    All the even and odd numbers between 1 and 100 are discussed here. What are the even numbers from 1 to 100? The even numbers from 1 to 100 are:

    Read More