Many word problems Involving unknown quantities can be translated for solving quadratic equations

Methods of solving quadratic equations are discussed here in the following steps.

**Step I:** Denote the unknown quantities by x, y etc.

**Step II:** use the conditions of the problem to establish in
unknown quantities.

**Step III:** Use the equations to establish one quadratic equation
in one unknown.

**Step IV:** Solve this equation to obtain the value of the
unknown in the set to which it belongs.

**Now we will learn how to frame the equations from word problem:**

**1.** The product of two consecutive integers is 132. Frame an equation for the statement. What is the degree of the equation?

**Solution:**

**Method
I:** Using only one unknown

Let the two consecutive integers be x and x + 1

Form the equation, the product of x and x + 1 is 132.

Therefore, x(x + 1) = 132

⟹ x\(^{2}\) + x - 132 = 0, which is quadratic in x.

This is the equation of the statement, x denoting the smaller integer.

** **

**Method
II:** Using more than one unknown

Let the consecutive integers be x and y, x being the smaller integer.

As consecutive integers differ by 1, y - x = 1 ........................................... (i)

Again, from the question, the product of x and y is 132.

So, xy = 132 ........................................... (ii)

From (i), y = 1 + x.

Putting y = 1 + x in (ii),

x(1 + x) = 132

⟹ x\(^{2}\) + x - 132 = 0, which is quadratic in x.

Solving the quadratic equation, we get the value of x. Then the value of y can be determined by substituting the value of x in y = 1 + x.

**2.** The length of a rectangle is greater than its breadth by
3m. If its area be 10 sq. m, find the perimeter.

**Solution:**

Suppose, the breadth of the rectangle = x m.

Therefore, length of the rectangle = (x + 3) m.

So, area = (x + 3)x sq. m

Hence, by the condition of the problem

(x + 3)x = 10

⟹ x\(^{2}\) + 3x - 10 = 0

⟹ (x + 5)(x - 2) = 0

So, x = -5,2

But x = - 5 is not acceptable, since breadth cannot be negative.

Therefore x = 2

Hence, breadth = 2 m

and length = 5 m

Therefore, Perimeter = 2(2 + 5) m = 14 m.

x = -5 does not satisfy the conditions of the problem length or breadth can never be negative. Such a root is called an extraneous root. In solving a problem, each root of the quadratic equation is to be verified whether it satisfies the conditions of the given problem. An extraneous root is to be rejected.

**Quadratic Equation**

**Introduction to Quadratic Equation**

**Formation of Quadratic Equation in One Variable**

**General Properties of Quadratic Equation**

**Methods of Solving Quadratic Equations**

**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**

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