A second degree polynomial is generally expressed as below:
P(x) = a ∙ x2 + b ∙ x2 + c, and a ≠ 0
P(x) can also be rewritten as: a(x - x1)(x - x2)
For any second degree polynomial that satisfies the conditions above we have:
x1 + x2 = - b/a
x1 ∙ x2 = c/a
x1 and x2 are the possible solutions for P(x)
The solutions of a second degree can be easily calculated using the quadratic formulas shown below:
x1 = (-b + √(b2 - 4ac)) / 2a
x2 = (-b - √(b2 - 4ac)) / 2a
b2 - 4ac is called the discriminant of the quadratic formula. By analyzing the discriminant it is possible to find out how many solutions P(x) has:
x1 = x2, if b2 - 4ac = 0, there exists only 1 solution
x1 ≠ x2, if b2 - 4ac > 0, there exists 2 solutions
there exists no solutions if b2 - 4ac < 0
Example 1: Predict the factors for the second degree polynomial equation x2-44x+ 435 = 0.
The given second degree polynomial equation is x2-44x+ 435 = 0.
Step 1: x2-44x+ 435 = x2- 29x- 15x+ (-29 x -15)
Step 2: x2-44x+ 435 = x(x- 29) - 15(x- 29)
Step 3: x2-44x+ 435 = (x- 29) (x- 15)
Step 4: x- 29 = 0 and x-15 = 0
Step 5: (x- 29) (x-15) = 0
The factors for the given second degree polynomial equation x2-44x+ 435 = 0 are therefore (x -29) and (x- 15).
Example 2: Find the roots of 3 x2 + x + 6.
In this example we will use the quadratic formula to determine its roots, where we have:
a = 3
b = 1
c = 6
since b2 - 4ac = 1 – 4*3*6 = -71 < 0, we conclude that the polynomial has no real roots but there are two complex roots, namely x = ( -1 + sqrt(71)i ) / 6 and x = ( -1 + sqrt(71)i ) / 6.
Example 3: Find the quadratic equation whose roots are 3, -2.
The given roots are 3, -2.
Sum of the roots = 3 + (-2) = 3 – 2 = 1;
Product of the roots = 3 x (-2) = -6.
We know the Quadratic Equation whose roots are given is
x2 – (sum of the roots) x + (product of the roots) = 0.
So, the required equation is x – (1) x + (-6) = 0.
ie. x – x – 6 = 0.