** Introduction:**

A quadratic function is a polynomial function of the type f(x) = ax^{2}+bx+c, where a ≠zero. The expression ax^{2} + bx + c mean that the polynomial that having 2 is the highest degree called as Quadratic function . By equating the quadratic function to zero, we get the quadratic equation. Let us see how to solve the quadratic word problems in the following section.

## Example 1:

** **The product of two consecutive negative integer is 12. What are the numbers?

**Solution:**

Remember that successive integers are one unit apart, so we take the numbers are x and x + 1. Multiplying to get the product

x (x + 1) = 12

x^{2} + x = 12

x ^{2} + x - 12 = 0

The above equation is in the form of quadratic equation. Factorize the above equation.

x^{2 }+ x -12 = 0

x^{2}+ 4x - 3x - 12 = 0

Take x as common from first two terms so the equation would be

x (x+4) - 3x - 12 = 0.

Now take -3 as common from last two terms

x (x+4) - 3 (x+4) = 0

(x - 3) (x + 4) = 0

x - 3 =0 and x+4 = 0

x = 3 and x = - 4

The solutions are x = 3 and x = -4. here we required to find the negative value, so ignore "x = 3" and

take x = –4. Then the other number is x + 1 = (–4) + 1 = –3.

The two negative numbers are –3 and –4.

## Example 2:

The length of a rectangular garden is 5 more than the width. The area is 25 sq. units. Form the quadratic equation.

**Solution:**

Let x be the width of the rectangular garden.

So, Length = x+5

The formula to find the length of the rectangular garden is = L * W

x * (x+5) = 25

x^{2} + 5x = 25

x^{2} +5x -25 =0

**Example 3:**

The product of two positive consecutive even integers is 48.Find the integers.

**Solution:**

Rememberthat successive even integers are two unit apart, so we take the numbers are x and x + 2. Multiplying to get the product

x (x + 2) = 48

x^{2} + 2x = 48

x ^{2} + 2x - 48="0

The above equation is in the form of quadratic equation. Factorize the above equation.

x^{2 }+2 x -48 = 0

x^{2}- 6x + 8x - 48 = 0

Take x as common from first two terms so the equation would be

x (x-6) + 8(x -6) = 0.

(x + 8) (x -6) = 0

x + 8 = 0 and x - 6 = 0

x = - 8 and x = 6

Thesolutions are x = - 8 and x = 6. Here we required to find the positive value, so ignore x =- 8 and take x = 6. Then the other number is x + 2 =6 + 2 = 8.

The two positive integers are 6,8.