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

2 Answers
Oct 10, 2017

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.

Oct 10, 2017

-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