Showing posts with label Quadratic equations. Show all posts
Showing posts with label Quadratic equations. Show all posts

Monday, November 19, 2012

Word Problems on Quadratic Equations

Introduction: How do you solve quadratic equation word problems
In algebra, a quadratic equation is a polynomial equation of the second degree. The general form is

ax² +bx+c=0

Where x stand for a variable, and a, b, and c, constants, with a ≠ 0. (If a = 0, the equation becomes a linear equation.)

The term "quadratic" comes from quadratics; Quadratic equations can be solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula.

How to Solve Word Problems:

We can solve the word problems by using quadratic equation that is write the problems in quadratic form and find out the result by factorization method or using quadratic formula. In math, the word problems are used to test the student's thinking knowledge. Some example problems for quadratic equation word problems. some times the quadratic formual is used to sovle the word problems.

Here we will explain how do you solve quadratic equation word problems

Example1:

How do you solve the width of a rectangle is 4 inches more than its length.  The area of the rectangle is 32 square inches. Find the dimensions of the rectangle.

Solution :

The formula for the rectangle area is LW = A

Step 1 - Draw the picture of the Rectangle.



Since the width is 4more than the length we let the rectangle        length = x and the width = x + 4

Step 2 - Write the equation using the   x(x +4) = 32

Step 3 - Solve the equation                    x2 +4x = 32
x2 + 4x.-32= 0
(x+8)(x-4)=0

X+8=0       |            x-4=0

X=8           |               x=4 the negative value is not established .so, we take x=8

The length of the rectangle is 8inch and width is 12 inch.

Examples

Example2:

How do you solve the width of a rectangle is 6 inches more than its length.  The area of the rectangle is 72 square inches. Find the dimensions of the rectangle.Having problem with how to solve a system of equations keep reading my upcoming posts, i will try to help you.

Solution :

The formula for the area of rectangle = LW = A

Step 1 - Draw the picture of the rectangle

Since the width is 4more than the length we let the rectangle        length = x and the width = x + 6

Step 2 - Write the equation using the   x(x +6) = 72

Step 3 - Solve the equation                    x2 +6x = 72
x2 + 6x.-72= 0

x2   +12x - 6x -72=0
(x-6)(x+12)=0

X-6=0       |            x+12=0

X=6           |               x=-12 the negative value is not accepted .so, we take x=6

The length of the rectangle is 6inch and width is 12 inch.


Example 3: The rectangle has 4 inches more than its width and the rectangle area is 90. Find out the length and width of rectangle.

Solution:

Step 1:

The dimensions of rectangle are width = x and length = x + 4.

Step 2:

Rectangle area formula is lw.

A = x(x+4) = 90

Step 3:

The quadratic equation is x2 + 4x – 32 = 0.

Solve the equation,

X2 - 4x + 8x – 32 = 0

X(x – 4) + 8(x – 4) = 0

(x – 4) (x + 8) = 0

X = 4, x = -8.

Therefore, the width of rectangle is 4 and the length of 4 and the width is 8.

Example 4 : Find two positive consecutive odd integers whose product is 63.

Solution:  

Let x be the first integer. Then the next odd integer is x + 2. So we have

x(x+2) = 63.

To solve this equation first we distribute and then set one side to zero. We have

x2+2x =63

x2+2x - 63 = 0

Factoring the quadratic equation x2 + 9x - 7x - 63 = 0

x(x + 9) - 7(x + 9) =0

(x + 9)(x - 7) = 0.

So the solutions are x = -9 or x = 7 Since we are looking for positive integers, we take x = 7

Hence  the numbers  are 7 and 9.

Example 5 :The triangle has three legs. The first triangle leg is 10 inches when the hypotenuse is 12 inches. Find out another leg.

Solution:

Take the unknown leg as x and use the Pythagorean Theorem.

x2 + 102 = 122

Simplify the equation,

x2 + 100 = 144

x2 = 144 – 100

x2 = 44

x = `sqrt(44)`

x = 6.63

Therefore, another leg measurement is 6.63 inches.

The quadratic equations are mainly used in real-time problems.

Wednesday, June 16, 2010

Introduction to quadratic equations

let me explain about quadratic equations,
The simplest relationship between two variables is when they are equal. The next simplest is when one variable is equal to the other multiplied by a constant. This relationship is discussed in the notebook on linear equations. In either case, the relationship is said to be "linear".

Another common sort of relationship between two variables is a "quadratic" relationship. A quadratic relationship involves the power 2, i.e. the operation of "squaring" somewhere within it.

An example of a quadratic term is x2, so is 3t2 and -.05q2. But the expressions x3 and x-2 are not quadratic, as the power involved is not 2.

A quadratic equation, then, is an equation that has a quadratic expression within it. For example: 4x2-5=0 or y=2x2. In the first case there is only one variable in the equation, in the second the equation gives the relationship between two variables.

Similarly a quadratic function is a function which involves the action of squaring somewhere within it. For example: f(x)=3x2+100.

Hope the above explanation was helpful.