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

Algebra/Linear Algebra

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

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