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

Question

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

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

Answer 3

Let's assume the two roots of the quadratic equation 𝑥^2 − 𝑏𝑥 + 𝑐 = 0 are 𝑝 and 𝑝+2, where 𝑝 is an even number.

We know that the sum of the roots of a quadratic equation is given by the formula:

𝑝 + (𝑝+2) = -𝑏

Simplifying this equation, we get:

2𝑝 + 2 = -𝑏

Now, let's multiply the two roots together, which gives us:

𝑝(𝑝+2) = 𝑝^2 + 2𝑝 = 𝑐

So, we have two equations:

2𝑝 + 2 = -𝑏 (Equation 1)
𝑝^2 + 2𝑝 = 𝑐 (Equation 2)

We can solve these equations simultaneously to find the values of 𝑏 and 𝑐.

Let's solve Equation 1 for 𝑏:

2𝑝 + 2 = -𝑏
𝑏 = -2𝑝 - 2

Now, substitute 𝑏 in Equation 2 with the value we found:

𝑝^2 + 2𝑝 = 𝑐

To find the value of 𝑐, we need to find 𝑝. Since 𝑝 is an even number, we can assume 𝑝 = 2𝑘, where 𝑘 is an integer.

Substituting 𝑝 = 2𝑘 in the equation:

(2𝑘)^2 + 2(2𝑘) = 𝑐
4𝑘^2 + 4𝑘 = 𝑐

Now, substitute 𝑏 = -2𝑝 - 2 and 𝑐 = 4𝑘^2 + 4𝑘 in Equation 𝑏^2 - 4𝑐 to find the exact value:

(-2𝑝 - 2)^2 - 4(4𝑘^2 + 4𝑘)

Simplifying this expression, we get the exact value of 𝑏^2 - 4𝑐.

Answer 4

To find the value of 𝑏2 − 4𝑐, we need to first determine the values of 𝑏 and 𝑐.

Given that the roots of 𝑥2 − 𝑏𝑥 + 𝑐 = 0 are two consecutive even numbers, we can assume the roots to be 𝑟 and 𝑟 + 2.

Using the sum and product of roots formula, we know that the sum of the roots (𝑟 + 𝑟 + 2) is equal to -𝑏 and the product of the roots is equal to 𝑐.

So, we have the following equations:

𝑟 + 𝑟 + 2 = -𝑏 ...(Equation 1)
𝑟 × (𝑟 + 2) = 𝑐 ...(Equation 2)

To solve these equations, we'll need either 𝑏 or 𝑐 to proceed further. However, we can simplify the expression 𝑏2 − 4𝑐 by substituting 𝑏 and 𝑐 in terms of 𝑟.

From Equation 1, we have:
𝑟 + 𝑟 + 2 = -𝑏
2𝑟 + 2 = -𝑏
𝑏 = -2𝑟 - 2

From Equation 2, we have:
𝑟 × (𝑟 + 2) = 𝑐
𝑟² + 2𝑟 = 𝑐

Now, we can substitute the values of 𝑏 and 𝑐 in 𝑏2 − 4𝑐:

𝑏2 − 4𝑐 = (-2𝑟 - 2)² - 4(𝑟² + 2𝑟)
= 4𝑟² + 8𝑟 + 4 - 4𝑟² - 8𝑟
= 0

Therefore, the exact value of 𝑏2 − 4𝑐 is 0.