Question

One integer is 3 more than 5 times another. The product is 54, what are the two integers?

Answer 1

The two integers are 18 and 3.

Explanation:

Consider the provided information.

It is given that One integer is 3 more than 5 times another.

Let us consider the integer is x.

Then 3 more than 5 times of x is:

`3+5x`

The product of the provided two numbers is 54. i.e.

`x(3+5x)=54`

`3x+5x^2=54`

`5x^2+3x-54=0`

Use the formula of the quadratic equation:

For the equation
`ax^2+bx+c=0` the solutions are:

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

By comparing the obtained equation with the above equation, it can be concluded that a = 5, b = 3 and c = -54. Substitute the these values in the above formula.

`x_(1,\:2)=(-3\pm √(3^2-4\cdot \:5\left(-54\right)))/(2\cdot \:5)`

`x_(1,\:2)=(-3\pm√(3^2+4\cdot \:5\cdot \:54))/(2\cdot \:5)`

`x_(1,\:2)=(-3\pm√(1089))/(10)`

`x_(1,\:2)=(-3\pm33)/(10)`

`x_1=(-3+33)/(10)` or
`x_2=(-3-33)/(10)`

`x_1=(30)/(10)` or
`x_2=(-36)/(10)`

`x_1=3` or
`x_2=(-18)/(5)`

As we have given that numbers are integer, therefore ignore
`x_2=(-18)/(5)`

Thus, the value of x is 3 or one number is 3.

Then 3 more than 5 times of x is: 3 + 5(3) = 18.

Therefore, the two integers are 18 and 3.