Forming a Quadratic
Polynomial by Knowing its Roots
Introduction
We are given two roots of a quadratic equation x = A and x =
B and asked to construct a quadratic polynomial. Believe it or not, this is (almost)
enough information to accomplish this task.
Since A and B are roots, that means:
0 = (x – A)*(x – B)
It’s now just a matter of some algebra:
0 = x^2 – A*x – B*x + A*B
0 = x^2 – (A+B)*x + A*B
Define this as the polynomial p(x) = x^2 – (A+B)*x + A*B
Example: A quadratic
polynomial p(x) with roots at x = 1 and x = 5.
A result is:
p(x) = x^2 – 6*x + 5
 |
| x^2 - 6*x + 5. All screen shots are generated from the HP Prime Emulator |
Note that this polynomial is concave up for all x in the real
numbers. In calculus, a function is concave
up when the second derivative is positive.
We can imagine bucket holding water.
Another Quadratic
Polynomial?
Yes, observe:
0 = (x – A)*(x – B)
0 = x^2 – (A+B)*x + A*B
Now multiply both sides by -1:
0 = -x^2 + (A+B)*x – A*B
Note that 0 * -1 = 0.
Let’s name this quadratic polynomial q(x) = -x^2 + (A+B)*x –
A*B.
Going back to our example, with roots x = 1 and x = 5, q(x)
is defined as:
q(x) = -x^2 + 6*x – 5
 |
| -x^2 + 6*x - 5 |
In this case this polynomial is concave down for all x. It’s like the bucket has been turned upside
down and water is spilling.
Can there be any
Other Polynomials?
0 = (x – A)*(x – B)
Multiply by sides by an amplifying factor C:
C * 0 = C * (x – A) * (x – B)
0 = C * (x^2 – (A + B)*x + A*B)
0 = C * x^2 – C*(A + B)*x + A*B*C
Name this polynomial r(x) = C * x^2 – C*(A + B)*x + A*B*C
Back to the example, where roots are located at x = 1 and x
= 5, let’s assume an amplifying factor of C = 3. The result is:
r(x) = 3*x^2 – 18*x + 15
 |
| 3*x^2 - 18*x + 15 |
The following is a set of four quadratic polynomials that
can be formed with roots x = 1 and x = 5:
 |
| Four quadratic polynomials with roots x = 1 and x = 5 |
|
Polynomial
|
Values
|
Color (see
above)
|
|
p(x) = x^2 – 6*x
+ 5
|
A = 1, B = 5, C = 1;
concave up
|
Blue
|
|
q(x) = -x^2 + 6*x
– 5
|
A = 1, B = 5, C = -1;
concave down
|
Red
|
|
r(x) = 3*x^2 –
18*x + 15
|
A = 1, B = 5, C = 3;
concave up
|
Green
|
|
s(x) =
-1/2*x^2 +
3*x – 5/2
|
A = 1, B = 5, C = -1/2;
concave down
|
Orange
|
Summary
Given the roots of the quadratic polynomial x = A and x = B,
with its amplifying factor C, possible quadratic polynomials can be formed by:
f(x) = C * x^2 – C*(A + B)*x + A*B*C
If C = 1, this simplifies to: f(x) = x^2 – (A + B)*x + A*B
If C = -1, this simplifies to: f(x) = -x^2 + (A + B)*x - A*B
If C>0, the second derivative is positive (f’’(x) = 2*C)
and the polynomial is concave up.
If C<0, the second derivative is negative (f’’(x) = -2*C)
and the polynomial is concave down.
Eddie
This blog is property of Edward Shore, 2018.