Question

A) show that x^2-8x+20 can be written in the form (x-a)^2+a where a is an integer

Answer 1

Answer:

a)

$x^2-8x+20$

$x^2-8x+16-16+20$

Completing the square

$(x-4)^2+4$

$(x-a)^2+a$

$a=4$

b)

You can prove that $x^2-8x+20$ is always positive, stating;

$x^2-8x+20=(x-4)^2+4=(x-4)^2+2^2$

The sum of two squared numbers can't be negative. Hence is always positive.

Also, if we take the discriminant of $x^2-8x+20$

$\Delta= \left(-8\right)^2-4\cdot \:1\cdot \:20\\\Delta= -16$

Once the discriminant is negative, the quadratic has no real root. So, the function never touches the real axis and the function lies above or below the real axis. If we take x as 0, y = 20, therefore, the function lies above the real axis. It is always positive.