Question

A positive real number is 2 less than another. When 4 times the larger is added to the square of the smaller, the result is 49. Find the numbers?

Answer 1

x=`-2+3sqrt5`
y=`3sqrt5`

Explanation:

Positive real number = `x`
Other number = `y`
First statement: A positive real number is 2 less than another.
Therefore `y=x+2`

Second statement: When 4 times the larger is added to the square of the smaller, the result is 49.
Since the larger number is `y`, we know that "4 times the
larger" means `4y`.
The square of the smaller would mean `x^2`.
If you put them together, it would look like `4y+x^2=49`.

Now create a system of equations like the following:
`y=x+2`
`4y+x^2=49`.

From the first equation, we can understand the value of `y` in terms of `x`, which is `x+2`.
Replace the `y` in the second equation with `x+2`.
`4(x+2)+x^2=49`
Now that we have an equation with only one variable, solve for `x`.
Below is a potential way one can solve this equation:
1. Expand the bracket.
`4x+8+x^2=49`
2. Form a quadratic equation (`ax^2+bx+c=0`)
`x^2+4x-41=0`
3, Use `(-b+-sqrt(b^2-4ac))/2`
(`a` is 1, `b` is 4, `c` is -41.)
`(-4+-sqrt(4^2-4*(-41)))/2`
`(-4+-(180)^(1/2))/2`
`-2+-3sqrt5`

Since the answer must be a positive real number, `x` will be `-2+3sqrt5`
If you place `-2+3sqrt5` into `y=x+2` :
`y=-2+3sqrt5+2`
You will get `3sqrt5` for `y`.