Question

Which two consecutive integers are such that the smaller added to the square of the larger is `21`?

Answer 1

None!

Explanation:

Let the larger no. be `x`.
Then, the smaller no. will be `x-1`.

According to the que,
`x^2+(x-1)= 21`
`=x^2+x-22=0`

Use quadratic formula with `a=1, b=1, c=-22`

`x=(−b+-sqrt(b^2−4ac))/(2a)`

`x=(−(1)+-sqrt((1)^2−4(1)(−22)))/(2(1))`

`x=(−1+-sqrt(89))/2`

So, there is no integer root for this equation.

Answer 2

`-5, -4`

Explanation:

Let n be the larger integer then: n - 1 is the smaller integer we have:

`n + (n - 1)^2 = 21`
`n + n^2 - 2n + 1=21`
`n^2-n-20=0`
`(n+4)(n-5)=0`
`n=-4,n=5`
`n-1=-5,n-1=4`
reject the positive roots thus:
-5 and -4 are the integers