Book:- ML aggarwal, Avichal publication, class10th, quadratic equation in one variable, chapter 5, exercise 5.5
This is an important ques and asked in previous year question paper
(i) If the product of two positive consecutive even integers is 288, find the integers.
(ii) If the product of two consecutive even integers is 224, find the integers.
(iii) Find two consecutive even natural numbers such that the sum of their squares is 340.
(iv) Find two consecutive odd integers such that the sum of their squares is 394.
Question no.2 , ML Aggarwal, chapter 5, exercise 5.5, quadratic equation in one variable, ICSE board,
Solution:
(i) If the product of two positive consecutive even integers is 288, find the integers.
Let us consider first positive even integer number be ‘2x’
Second even integer number be ‘2x + 2’
So according to the question,
2x × (2x + 2) = 288
4x2 + 4x – 288 = 0
Divide by 4, we get
x2 + x – 72 = 0
Let us factorize,
x2 + 9x – 8x – 72 = 0
x(x + 9) – 8(x + 9) = 0
(x + 9) (x – 8) = 0
So,
(x + 9) = 0 or (x – 8) = 0
x = -9 or x = 8
∴ Value of x = 8 [since, -9 is not positive]
First even integer = 2x = 2 (8) = 16
Second even integer = 2x + 2 = 2(8) + 2 = 18
(ii) If the product of two consecutive even integers is 224, find the integers.
Let us consider first positive even integer number be ‘2x’
Second even integer number be ‘2x + 2’
So according to the question,
2x × (2x + 2) = 224
4x2 + 4x – 224 = 0
Divide by 4, we get
x2 + x – 56 = 0
Let us factorize,
x2 + 8x – 7x – 56 = 0
x(x + 8) – 7(x + 8) = 0
(x + 8) (x – 7) = 0
So,
(x + 8) = 0 or (x – 7) = 0
x = -8 or x = 7
∴ Value of x = 7 [since, -8 is not positive]
First even integer = 2x = 2 (7) = 14
Second even integer = 2x + 2 = 2(7) + 2 = 16
(iii) Find two consecutive even natural numbers such that the sum of their squares is 340.
Let us consider first positive even natural number be ‘2x’
Second even number be ‘2x + 2’
So according to the question,
(2x)2 + (2x + 2)2 = 340
4x2 + 4x2 + 8x + 4 – 340 = 0
8x2 + 8x – 336 = 0
Divide by 8, we get
x2 + x – 42 = 0
Let us factorize,
x2 + 7x – 6x – 56 = 0
x(x + 7) – 6(x + 7) = 0
(x + 7) (x – 6) = 0
So,
(x + 7) = 0 or (x – 6) = 0
x = -7 or x = 6
∴ Value of x = 6 [since, -7 is not positive]
First even natural number = 2x = 2 (6) = 12
Second even natural number = 2x + 2 = 2(6) + 2 = 14
(iv) Find two consecutive odd integers such that the sum of their squares is 394.
Let us consider first odd integer number be ‘2x + 1’
Second odd integer number be ‘2x + 3’
So according to the question,
(2x + 1)2 + (2x + 3)2 = 394
4x2 + 4x + 1 + 4x2 + 12x + 9 – 394 = 0
8x2 + 16x – 384 = 0
Divide by 8, we get
x2 + 2x – 48 = 0
Let us factorize,
x2 + 8x – 6x – 48 = 0
x(x + 8) – 6(x + 8) = 0
(x + 8) (x – 6) = 0
So,
(x + 8) = 0 or (x – 6) = 0
x = -8 or x = 6
When x = -8, then
First odd integer = 2x + 1 = 2 (-8) + 1 = -16 + 1 = -15
Second odd integer = 2x + 3 = 2(-8) + 3 = -16 + 3 = -13
When x = 6, then
First odd integer = 2x + 1 = 2 (6) + 1 = 12 + 1 = 13
Second odd integer = 2x + 3 = 2(6) + 3 = 12 + 3 = 15
∴ The required odd integers are -15, -13, 13, 15.