Show that the root equation (x-a)(x-b)=k^2 are always real numbers
x^2 - (a+b) x + a b - k^2 = 0
x = [(a+b) +/- sqrt { (a+b)^2 -4(ab-k^2)} ]/2
if
(a+b)^2 -4(ab-k^2)
is positive, there is no complexity
= a^2 + 2 ab +b^2 -4ab + 4k^2
= a^2 -2ab + b^2 + 4 k^2
= (a-b)^2 + 4 k^2
BOTH of those terms are positive being squares of real numbers, whew we did it !
show that the root of the equation x^2+2x=2x(2a+2b+1)(2a+2b-1) are integers if a and b are integer
To show that the root equation (x-a)(x-b) = k^2 always yields real numbers, we can consider the discriminant of the quadratic equation. The discriminant, denoted as Δ, is a value that can be derived from the coefficients of a polynomial equation and can be used to determine the nature of its roots.
In this case, we are considering the quadratic equation (x-a)(x-b) = k^2. Let's expand it and simplify:
x^2 - (a+b)x + ab = k^2
Now, the discriminant Δ is calculated using the formula:
Δ = (b^2 - 4ac)
For our equation, the coefficients a, b, and c are:
a = 1
b = -(a + b) = -a - b
c = ab - k^2
Substituting these values into the discriminant formula, we have:
Δ = (-a - b)^2 - 4(1)(ab - k^2)
Expanding and simplifying further:
Δ = a^2 + b^2 + 2ab - 4ab + 4k^2
Δ = a^2 + b^2 - 2ab + 4k^2
We can rewrite this equation as:
Δ = (a - b)^2 + 4k^2
Since k^2 is always non-negative (k^2 ≥ 0), then the discriminant Δ will always be non-negative as well. This means that Δ ≥ 0, indicating that the roots of the equation are always real numbers.
Therefore, we have shown that the root equation (x-a)(x-b) = k^2 always yields real numbers.