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.
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