Question

The product of two consecutive integers is zero, find the integers. Show work

Answer 1

Answer:

0,1

Step-by-step explanation:

Let n represent the first interger. and n+1 represent the next interger.

$n \times (n + 1) = 0$

${n}^{2} + n = 0$

Complete the square.

$( \frac{n}{2} {)}^{2} = \frac{ {n}^{2} }{4}$

${n}^{2} +n + \frac{ {n}^{2} }{4} = \frac{ {n}^{2} }{4}$

$(n + \frac{n}{2} ) {}^{2} = \frac{n {}^{2} }{4}$

Take the square root.

$(n + \frac{n}{2} ) = \frac{n}{2}$

$n = 0$

Plug 0 in for 0+1.

$0 + 1 = 1$

So the consective intergers are 0,1.