Find two consecutive odd integers such that the square of the first added to 3 times the second, is 24. Part a: Define the variables. Part b: Set up an equations that can be solved to find the integers. Part c: Find the integers.

Respuesta :

Given:

There are two consecutive odd integers such that the square of the first added to 3 times the second, is 24.

To find:

Part a: Define the variables.

Part b: Set up an equations that can be solved to find the integers.

Part c: Find the integers.

Solution:

Part a:

Let x be the first odd integers. Then next consecutive odd integer is [tex]x+2[/tex], because the difference between two consecutive odd integers is 2.

Part b:

Square of first odd integers = [tex]x^2[/tex]

Three times of second odd integers = [tex]3(x+2)[/tex]

It is given that the sum of square of first odd integers and three times of second odd integers is 24. So, the required equation is:

[tex]x^2+3(x+2)=24[/tex]

Part c:

The equation is:

[tex]x^2+3(x+2)=24[/tex]

It can be written as:

[tex]x^2+3x+6=24[/tex]

[tex]x^2+3x+6-24=0[/tex]

[tex]x^2+3x-18=0[/tex]

Splitting the middle term, we get

[tex]x^2+6x-3x-18=0[/tex]

[tex]x(x+6)-3(x+6)=0[/tex]

[tex](x-3)(x+6)=0[/tex]

[tex]x=3,-6[/tex]

-6 is not an odd integer, so [tex]x=3[/tex] and the first odd integer is 3.

Second odd integer = [tex]x+2[/tex]

                                  = [tex]3+2[/tex]

                                  = [tex]5[/tex]

Therefore, the two consecutive odd integers are 3 and 5.

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