Question

Three consecutive positive integer are such that the sum of the squarebof thevfirst and the product oftheir two is 46Find the integers

Answer 1

Given :

• Three consecutive positive integer are such that the sum of the square of the first and the product of their two integers is 46.

To find :

• The integers =?

Step-by-step explanation :

Let, the first consecutive positive integer be x.

Then, the second consecutive positive integer be x + 1.

And, the third consecutive positive integer be x + 2.

It is Given that,

The sum of the square of the first and the product of their two integers is 46.

According to the question :

x² + (x + 1)(x + 2) = 46

x² + x² + 3x + 2 = 46

2x² + 3x - 44 = 0

2x² - 8x + 11x - 44 = 0

2x(x - 4) + 11(x - 4) = 0

(2x + 11) (x - 4) = 0

Now,

Value of x,

2x + 11 = 0

2x = - 11

x = - 11/2 [Ignore Negative]

Or,

x - 4 = 0

x = 4.

Now,

x = - 11/2 Or x = 4.

We can't take the negative values. So, x = 4.

Therefore,

The first consecutive positive integer, x=4

Then, the second consecutive positive integer, x + 1 = 4 + 1 = 5

And, the third consecutive positive integer, x + 2 = 4 + 2 = 6

Answer 2

Question:

Three consecutive positive integer are such that the sum of the square of the first and the product of other two is 46 . Find the integers.

Answer:

The integers are : 4,5 and 6

Given:

►There are three consecutive positive integers.8

►The sum of the square of the first and the product of other two is 46.

To Find:

The Integers.

Solution:

We are given,

►There are three consecutive positive integers.

►The sum of the square of the first and the product of other two is 46.

Let, there consecutive positive integers are m , m+1 and m+2 .

By given condition,

${m}^{2} + (m + 1)(m + 2) = 46$

${m}^{2} + {m}^{2} + 2m + m + 2 = 46$

$2 {m}^{2} + 3m + 2 - 46 = 0$

$2 {m}^{2} + 3m - 44 = 0$

Now, We have to find factors of the equation.

Here,

a = 2 ; b = 3 ; c = -44

$m \: = \: \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a}$

$m \: = \: \frac{ - 3± \sqrt{ {3}^{2} - 4(2)( - 44) } }{2(2)}$

$m \: = \: \frac{ - 3± 19}{4}$

$m \: = \: \frac{ - 3 + 19}{4} \: and \: m \: = \: \frac{ - 3 - 19}{4}$

$m \: = \: \frac{ 16}{4} \: and \: m \: = \: \frac{ - 21}{4}$

$m \: = \: 4 \: and \: m \: = \: \frac{ - 11}{2}$

Here , We can see that $\rm{\frac{ -11}{2}}$ is not an integer.

Therefore, there is only one integer which is 4.

Therefore,

m+1 = 4+1 =5

m+2 =4+2=6

Therefore the integers are 4 , 5 and 6