Find two consecutive numbers who squares differ by 31 I got 2 and 31 but I got it wrong someone help me please

consecutive number are one apart

so let the first be x
then the 2nd is x+1

their squares differ by 1 ---> (x+1)^ - x^2 = 31
x^2 + 2x + 1 - x^2 = 31
2x = 30
x = 15

so the first is 15, the next one is 16

check:
16^2 - 15^2
= 256 - 225
= 31

all is good!

We know in January it was 11 degrees Celsius

In December it is -1 degree Celsius

Take that and put -1 on the left.
Now put a < sign next to the one.
After that put the 11 next to the < sign.Now you have your inequality of -1<11.

Write an inequality for the statement:The average temperature in January was 11° C which is warmer than Decembers average temperature of -1° C.

To find two consecutive numbers whose squares differ by 31, let's set up an equation based on the problem:

Let's assume that the first number is "x." The next consecutive number will be "x + 1."

According to the problem, the squares of these two numbers differ by 31. Therefore, we can write the equation as:

(x + 1)^2 - x^2 = 31

Expanding the equation, we get:

(x^2 + 2x + 1) - x^2 = 31

Simplifying further:

2x + 1 = 31

Now, let's solve for x:

2x = 31 - 1

2x = 30

Dividing both sides of the equation by 2:

x = 15

So, the first number is 15. Now, we can find the next consecutive number by adding 1:

x + 1 = 15 + 1 = 16

Therefore, the consecutive numbers whose squares differ by 31 are 15 and 16.