Find the exact value of 𝑏2 − 4𝑐 if the roots of 𝑥2 − 𝑏𝑥 + 𝑐 = 0 are two consecutive even

Question

numbers.

Answer 1

Let the two consecutive even numbers be $2n$ and $2n+2$. By Vieta's formulas, we know that the sum of the roots of the quadratic equation $x^2-bx+c=0$ is $b$ and the product of the roots is $c$. Thus, $$b=2n+2n+2=4n+2$$and $$c=2n\cdot (2n+2)=4n^2+4n.$$We want to find $b^2-4c$. We have $$b^2-4c=(4n+2)^2-4(4n^2+4n)=16n^2+16n+4-16n^2-16n=4.$$Therefore, $b^2-4c=\boxed{4}.$

Answer 2

Looks like the bot is writing in weird code again.
Here is the solution in a form that humans can read:

Let the two roots be m and m+2

The sum of roots = m + m+2 = 2m + 2
product of roots = m(m+2) = m^2 + 2m

from the equation: sum of roots = b
product of roots = c
so b^2 - 4x = (2m+2)^2 - 4(m^2 + 2m)
= 4m^2 + 8m + 4 - 4m^2 - 8m + 4
= 4