Question

find the three second degree polynomial​

Answer 1

can you please elaborate?

Answer 2

$\huge\fcolorbox{black}{lime}{Answer}$

A second degree polynomial, also referred as a quadratic equation can be expressed as below:

ax2 + bx + c = 0

to solve the equation we can use the quadratic formulas as shown below:

x1 = (-b + (b2-4ac)1/2)/2a

x2 = (-b - (b2-4ac)1/2)/2a

a quadratic equation has two solutions when b2-4ac > 0

a quadratic equation has only one solution when b2-4ac = 0

a quadratic equation has no solution when b2-4ac < 0

Example:-

2x2 + 6x + 1 = 0

b2-4ac = 62-4 x 2 x 1 = 28, since 28 > 0, we can conclude that there exists two solutions

x1 = (-b + (b2-4ac)1/2)/2a = -0.177

x2 = (-b - (b2-4ac)1/2)/2a = -2.822