Math Examples of Quadratic Equations

The zero of polynomial equation can be derived by equating P (y) to zero. ‘A’ is zero or root of polynomial P(X) if and only if P(X) = 0. For example,
Let P(X) = 5X³ − 4X² + 7X − 8.  Then one of the roots or zeros of this quadratic polynomial is 1. It can be proved by substituting the value y = 1 in the given polynomial.
P (1) = 5 (1)³  - 4 (1)² + 7 (1) – 8 = 5 – 4 + 7 – 8 = 0.
If (p + q i) is a root of the polynomial equation then (p – q i) is also a root of quadratic polynomial function.

Let’s consider an example in which if (2 + m) is a root to f (m) = - m 2 + 4 m – 5, then (2 – m) is also a root of the function.
Let’s put (2 + m) in the given equation
f (m) = - (2 + m) ² + 4 ( 2 + m) – 5,
        = - (4 – 1 + 4 m) + 3 + 4 m,
     = - 3 – 4 m + 3 + 4 m = 0,
f (m) = 0. If we substitute (2 – m) into the function, similar results will be obtained.

Example: If 5 is one zero of the quadratic polynomial x^2 minus kx minus 15 then value of k is?
Given polynomial equation is x² – k x – 15. This equation can be written in the form of a function as:
f (x) = x² – kx – 15,
In this problem, it is given to us that one zero exist at value of 'x' equals to 5. To prove this, substitute x = 5. This can be derived as:
f (5) = 5² – 5k – 15 = 0,
25 – 5k – 15 = 0,
10 – 5 k = 0,
5 k = 10 and hence k = 2.
Hence, if 5 is one zero of quadratic polynomial x² minus kx minus 15 then value of k is 2.