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Find two positive consecutive odd integers such that the square of the larger integer is one less than twice the square of the small integer.

Respuesta :

calculista
Let
x---------------> first  positive odd integer
x+2------------> second  positive consecutive odd integer

we know that
(x+2)²=2x²-1---------> x²+4x+4=2x²-1-------> x²-4x-5
x²-4x-5=0
using a graph tool--------> to calculate the quadratic equation
see the attached figure

the solution is
x=5

the answer is
the first  positive odd integer x is 5
the second  positive consecutive odd integer x+2 is 7
Ver imagen calculista

Answer:

smaller integer = 5

Larger integer = 7

Step-by-step explanation:

Let the first odd integer be x

Then the next consecutive integer will be x + 2

Now, The square of larger integer is 1 less than twice the square of the smaller integer.

As, x < x + 2

Smaller integer = x

Larger integer = x + 2

⇒ (x + 2)² + 1 = 2·x²

⇒ x² + 4 + 4x + 1 = 2x²

⇒ x² - 4x - 5 = 0

⇒ x² - 5x + x - 5 = 0

⇒ x(x - 5) + 1(x - 5) = 0

⇒ (x - 5)(x + 1) = 0

⇒ x = 5 or x = -1

But we need positive integer so x = -1 is rejected.

Hence, x = 5

So, smaller integer = 5

Larger integer = 7

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