SOLUTION: The product of two consecutive integers is -4 more than the square of the smaller one.
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Question 261679: The product of two consecutive integers is -4 more than the square of the smaller one.
Found 3 solutions by CharlesG2, mananth, texttutoring:
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
The product of two consecutive integers is -4 more than the square of the smaller one.
x smaller integer
x+1 next integer
x * (x+1) = x^2 - 4
x^2 + x = x^2 - 4
x = -4
x + 1 = -3
integers are -4 and -3
check:
-4 * -3 = 12
-4^2=16
16 - 4 = 12
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
The product of two consecutive integers is -4 more than the square of the smaller one.
Let one integer be x
The other will be x+1
X(x+1)= x^2-4
X^2+x=x^2-4
X=-4
& -3
Answer by texttutoring(324) (Show Source): You can put this solution on YOUR website!
Let x = the first integer and
let y = the second integer
The numbers are consecutive, which gives us Eqn 1: y = x+1
Their product is -4 more than the square of the smaller one (x).
This gives us Eqn. 2:
We will now insert y = x+1 from Eqn. 1 into Eqn. 2:
Move everything to the left side and collect like terms:
Which means that y = 5.
The two numbers are 4 and 5.
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