Question

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The product of two consecutive positive integers is 42. Write and solve a quadratic equation to find the solutions. Then show how you use those solutions to find the two consecutive positive integers.

Answer 1

$n(n+1)=42\\n^2+n-42=0\\n^2-6n+7n-42=0\\n(n-6)+7(n-6)=0\\(n+7)(n-6)=0\\n=-7 \vee n=6$

-7 is not positive

$n=6\\n+1=7$

The numbers are 6 an 7.

Answer 2

Answer:

$\boxed{\text{6 and 7}}$

Step-by-step explanation:

1. Set up the equation

        Let x = the first integer. Then

         x + 1 = the next integer

     x(x + 1) = the product of the integers

     x(x + 1) = 42

       x² + x = 42     Distributed the x

x² + x - 42 = 0

2. Solve for x

(x + 7)(x - 6) = 0            Factored the quadratic

x + 7 = 0     x - 6 = 0     Applied zero product rule

     x = -7         x = 6     Solved the binomials

We reject x = -7, because x must be positive

    x = 6

x + 1 = 7

$\text{The two consecutive positive integers are \boxed{\textbf{6 and 7}}}$

Check:

6(6+ 1) = 42

   6(7) = 42

     42 = 42

OK.