Question

the product of two consecutive integers is 420. an equation is written in standard form to solve for the smaller integer by factoring.
what is the constant of the quadratic expression in this equation?
$x^2 + x + \square = 0$

Explanation:

Step1: Define the variables

Let the smaller integer be \( x \). Then the next consecutive integer is \( x + 1 \).

Step2: Set up the equation

The product of the two consecutive integers is 420, so we have the equation \( x(x + 1)=420 \).

Step3: Expand the equation

Expanding the left - hand side, we get \( x^{2}+x = 420 \).

Step4: Rewrite in standard form

To write the quadratic equation in standard form \( ax^{2}+bx + c = 0 \), we subtract 420 from both sides of the equation \( x^{2}+x=420 \). So the equation becomes \( x^{2}+x - 420=0 \).

Answer:

• 420