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 .

For more details on consecutive numbers follow the link below:

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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

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.

1)is not similar since corresponding sides are not in the same ratio

2)are similar since corresponding sides are in the same ratio and corresponding angles are equal

hope it will

#faazmessi