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mixas84 [53]
3 years ago
8

La suma dels quadrats de tres nombres naturals consecutius i múltiples de 3 és igual a 450. Planteja una equació de segon grau i

calcula els tres nombres.
Mathematics
1 answer:
Firlakuza [10]3 years ago
7 0

Answer:

The 3 numbers are 9, 12, and 15.

Step-by-step explanation:

I will answer in English, below the answer you can see a translation in Catalan.

This can be translated to:

The sum of the squares of three consecutive and multiple natural numbers of 3 is equal to 450. Pose a quadratic equation and calculate the three numbers.

First, a multiple of 3 can be written as:

3*n

The consecutive multiple of 3 is:

3*n + 3

The consecutive multiple of 3 is:

3*n + 3 + 3

Then the sum of their squares is:

(3*n)^2 + (3*n + 3)^2 + (3*n + 6)^2

And we know that this is equal to 450, then we need to solve the equation:

(3*n)^2 + (3*n + 3)^2 + (3*n + 6)^2 = 450

let's solve this:

9*n^2 + (9*n^2 + 2*(3*n)*3 + 9) + (9*n^2 + 2*(3*n)*6 + 36) = 450

27*n^2 + 54*n + 45 = 450

we can write this as:

27*n^2 + 54*n + 45 - 450 = 0

27*n^2 + 54*n - 405 = 0

The solutions of this equation are given by the Bhaskara's formula, and the solutions are:

n = \frac{-54 \pm \sqrt{54^2 -4*27*(-415)}  }{2*27} = \frac{-54 \pm 216 }{54}

we know that our numbers are naturals, then the numbers are positives, which means that we only care for the positive solution of n, which is:

n = (-54 + 216)/54 = 3

Then the 3 numbers are:

3*n = 3*3 = 9

(3*n + 3) = (3*3 + 3) = 12

(3*n + 6) = (3*3 + 6) = 15

In Catalan:

En primer lloc, es pot escriure un múltiple de 3 com:

3 * n

El múltiple consecutiu de 3 és:

3 * n + 3

El múltiple consecutiu de 3 és:

3 * n + 3 + 3

Llavors, la suma dels seus quadrats és:

(3 * n) ^ 2 + (3 * n + 3) ^ 2 + (3 * n + 6) ^ 2

I sabem que això és igual a 450, llavors hem de resoldre l’equació:

(3 * n) ^ 2 + (3 * n + 3) ^ 2 + (3 * n + 6) ^ 2 = 450

resolem això:

9 * n ^ 2 + (9 * n ^ 2 + 2 * (3 * n) * 3 + 9) + (9 * n ^ 2 + 2 * (3 * n) * 6 + 36) = 450

27 * n ^ 2 + 54 * n + 45 = 450

podem escriure això com:

27 * n ^ 2 + 54 * n + 45 - 450 = 0

27 * n ^ 2 + 54 * n - 405 = 0

Les solucions d’aquesta equació vénen donades per la fórmula de Bhaskara, i les solucions són:

n = \frac{-54 \pm \sqrt{54^2 -4*27*(-415)}  }{2*27} = \frac{-54 \pm 216 }{54}

sabem que els nostres nombres són naturals, aleshores els nombres són positius, cosa que significa que només ens importa la solució positiva de n, que és:

n = (-54 + 216) / 54 = 3

Llavors els 3 nombres són:

3 * n = 3 * 3 = 9

(3 * n + 3) = (3 * 3 + 3) = 12

(3 * n + 6) = (3 * 3 + 6) = 15

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