Which two consecutive integers are such that the smaller added to the square of the larger is 2121?

2 Answers
Gaya3
Oct 10, 2017

None!

Explanation:

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

According to the que,
x^2+(x-1)= 21x2+(x−1)=21
=x^2+x-22=0=x2+x−22=0

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

x=(−b+-sqrt(b^2−4ac))/(2a)x=−b±√b2−4ac2a

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

x=(−1+-sqrt(89))/2x=−1±√892

So, there is no integer root for this equation.

iceman
Oct 10, 2017

-5, -4−5,−4

Explanation:

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

n + (n - 1)^2 = 21n+(n−1)2=21
n + n^2 - 2n + 1=21n+n2−2n+1=21
n^2-n-20=0n2−n−20=0
(n+4)(n-5)=0(n+4)(n−5)=0
n=-4,n=5n=−4,n=5
n-1=-5,n-1=4n−1=−5,n−1=4
reject the positive roots thus:
-5 and -4 are the integers

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