Find two consecutive integers whose product is 6 less than the square of the smaller number

The two consecutive integers whose product is 6 less than the square of the smaller number integers are and .

Let the two consecutive integers be and respectively.

Given, the product of both the integers is 6 less than the square of the smaller number.

So we can write it in equation form as,

On solving ,

Further,

Or,

Now the other integer be   or .

Hence the two consecutive integers are and .

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Hi there!

Your answer is:

The smaller integer is -6

The larger integer is -5

Step-by-step explanation:

Two consecutive integers:

x, x+1

Product= multiply

(x) × (x+1) = x^2 - 6

x^2 + x= x^2 -6

-x^2

x= -6

Plug back in

First integer (smaller):

-6

Second integer (larger):

-6+1

-5

Check your work!

(-6) × (-6+1)

(-6)× (-5)

30 = (-6)^2 -6

30 = 36-6

30= 30 ✓✓✓✓

That's how we determine that this is the correct answer

Hopefully this helps

step-by-step explanation: