by bibi
(city)
The sum of squares of two consecutive odd positive integers is 202. Find the integers.
Solution
Let 2x + 1 be the first odd integer
Then, 2x + 3 will be the second integer
Now square each odd integer
(2x + 1)^{2} = (2x + 1) × (2x + 1)
(2x + 1) × (2x + 1) = 2x × 2x + 2x × 1 + 1 × 2x + 1 × 1
(2x + 1) × (2x + 1) = 4x^{2} + 2x + 2x + 1 = 4x^{2} + 4x + 1
(2x + 3)^{2} = (2x + 3) × (2x + 3)
(2x + 3) × (2x + 3) = 2x × 2x + 2x × 3 + 3 × 2x + 3 × 3
(2x + 3) × (2x + 3) = 4x^{2} + 6x + 6x + 9 = 4x^{2} + 12x + 9
Since the sum of the squares equal 202, we get
4x^{2} + 4x + 1 + 4x^{2} + 12x + 9 = 202
8x^{2} + 16x + 10 = 202
8x^{2} + 16x + 10 - 202 = 202 - 202
8x^{2} + 16x - 192 = 0
Divide everything by 8
We get x^{2} + 2x - 24 = 0
To factor the left side of this equation, look for factors of -24 that will add up to 2.
6 × -4 = -24 and 6 + -4 = 2.
We get (x + 6) × (x + -4) = 0
We have two equation to solve
x + 6 = 0 and x + -4 = 0
x = -6 and x = 4
x = -6 cannot be used since the integers are positive
we can use x = 4.
The first integer is 2x + 1 = 2 × 4 + 1 = 9
The seond integer is 2x + 3 = 2 × 4 + 3 = 11
Indeed 9^{2} + 11^{2} = 81 + 121 = 202
Jul 06, 18 12:29 PM
Learn how to solve two types of logarithmic equations
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Jul 06, 18 12:29 PM
Learn how to solve two types of logarithmic equations
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