The ages of 2 children are 11 and 8 years in how many years time will the product of their ages be 208
(11+x)(8+x) = 208
solve in the usual way
There can be multiple answers for this.
But only one will make sense.
There are only two factors of 208 which differ by 3.
To find out in how many years' time the product of their ages will be 208, let's start by assigning variables to the ages of the two children. Let's say the ages of the children are x and y.
From the given information, we know that the age of one child is 11 and the age of the other child is 8. So we can write the following equations:
x = 11 (age of one child)
y = 8 (age of the other child)
We also know that we want to find out how many years' time, denoted by n, it will take for the product of their ages to be 208. So we can write the equation:
(x + n) * (y + n) = 208
Now we can substitute the values of x and y into the equation:
(11 + n) * (8 + n) = 208
Expanding and simplifying the equation, we get:
88 + 19n + n^2 = 208
Rearranging the equation to form a quadratic equation, we have:
n^2 + 19n + 88 = 208
Now, we can subtract 208 from both sides of the equation to get a quadratic equation equal to zero:
n^2 + 19n + 88 - 208 = 0
Simplifying further:
n^2 + 19n - 120 = 0
Now, we can solve this quadratic equation to find the value of n using factoring, completing the square, or the quadratic formula. Solving using factoring:
(n + 15)(n - 8) = 0
This equation is satisfied when either (n + 15) = 0 or (n - 8) = 0. So we have two possible values for n:
n + 15 = 0 -> n = -15
n - 8 = 0 -> n = 8
However, in this context, we are looking for a positive value for n, representing the number of years in the future. Therefore, the answer is:
In 8 years' time, the product of their ages will be 208.