Question

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.

Answer 1

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 $x+2$, because the difference between two consecutive odd integers is 2.

Part b:

Square of first odd integers = $x^2$

Three times of second odd integers = $3(x+2)$

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:

$x^2+3(x+2)=24$

Part c:

The equation is:

$x^2+3(x+2)=24$

It can be written as:

$x^2+3x+6=24$

$x^2+3x+6-24=0$

$x^2+3x-18=0$

Splitting the middle term, we get

$x^2+6x-3x-18=0$

$x(x+6)-3(x+6)=0$

$(x-3)(x+6)=0$

$x=3,-6$

-6 is not an odd integer, so $x=3$ and the first odd integer is 3.

Second odd integer = $x+2$

                                  = $3+2$

                                  = $5$

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