Page 24 - CHAPTER 4 (Quadratic equations)

CHAPTER 4

QUADRATIC EQUATIONS

Example 5: Find two consecutive odd positive integers, sum of whose

square is 290.

Solution : Let the smaller of the two consecutive odd positive integers

be x. Then, the second integer will be x+2 . According to the question.

x +(x+2) =290
2
2
2
i.e., x +x +4x+4=290
2

2
i.e., 2x +4x-286=0

i.e., x +2x-143=0
2

which is a quadratic equation in x.

using the quadratic formula, we get

x= -2±√4+572 = -2±√576 = -2±24
2 2 2

i.e., x=11 or x=-13

But x is given to be an odd positive integer. Therefore, x≠ −13, = 11.

Thus, the two consecutive odd integers are 11 and 13.

2
12
Check : 11 +13 =121+169=290
Example 6: A rectangular park is to be designed whose breadth is 3

m less than its length . Its area is to be 4 square metres more than the

area of a park that has a already been made in the shape of an

isosceles triangle with its base as the breadth of the rectangular park

and of altitude 12m. find its length and breadth.

Solution: Let the breadth of the reactangular park be xm.

So, its length =(x+3)m

2
2
2
Therefore , the area of the rectangular park=x(x+3)m =(x +3x)m .

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