PART A:
The given quadratic equation is 2x²-10x-8=0
The radicand is given by b²-4ac where a, b, and c are the constants in a quadratic form ax²+bx+c
From the given equation, we have
a = 2
b = -10
c = -8
Radicand b²-4ac = (-10)² - 4(2)(-8) = 100 + 64 = 164
The radicand is >0 hence the quadratic equation has two distinct roots
PART B:
4x²-12x+5 = 0
We can use the factorization method to solve the equation
Firstly, we multiply 4 by 5 to get 20
Then we find the pair of numbers that multiply gives 20 and sum gives -12
The pair of number is -2 and -10
Rewriting the equation
4x²-2x-10x+5 = 0
2x(2x-1)-5(2x-1) = 0
(2x-1)(2x-5) = 0
2x-1 = 0 and 2x-5 = 0
x = 1/2 and x = 5/2
