A monic quadratic is a quadratic in which the coefficient of the quadratic term is 1. For example, r^2 - 3r + 7 is a monic quadratic, but 3t^2 - 3t + 1 is not.
A teacher writes a monic quadratic on the board.
Joanie copies the quadratic onto her paper, but writes down the wrong constant term (but the correct quadratic and linear terms). She correctly factors the quadratic that she wrote down on her paper, and determines that her quadratic has roots -16 and 2.
Kelvin is factoring the same quadratic that the teacher wrote on the board. He also copies the quadratic onto his paper, but he writes down the wrong coefficient for the linear term (but the correct quadratic and constant terms). He correctly factors the quadratic that he wrote down, and determines that his quadratic has roots -36 and 2.
What are the roots of the quadratic that the teacher wrote on the board?
(x+16)(x-2) = x^2 + 14x - 32
(x+36)(x-2) = x^2 + 34x - 72
we know that +14 is correct, but -32 is wrong
we know that +34 is wrong, but -72 is correct
So, the desired quadratic must be x^2 + 14x - 72 = (x+18)(x-4)
To find the roots of the quadratic that the teacher wrote on the board, let's first understand how Joanie and Kelvin factored their versions of the quadratic.
Joanie's quadratic has roots -16 and 2. This means that the factors of her quadratic can be written as (x + 16) and (x - 2). Multiplying these factors together should give us her quadratic.
So, (x + 16)(x - 2) = x^2 - 2x + 16x - 32 = x^2 + 14x - 32. This is the quadratic that Joanie wrote down.
Similarly, Kelvin's quadratic has roots -36 and 2. Using the same logic, we can write the factors of his quadratic as (x + 36) and (x - 2). Multiplying these factors together should give us his quadratic.
So, (x + 36)(x - 2) = x^2 - 2x + 36x - 72 = x^2 + 34x - 72. This is the quadratic that Kelvin wrote down.
Now, let's compare the linear terms of their quadratics: -2x for Joanie and 34x for Kelvin. We know that the quadratic the teacher wrote on the board has a monic form, so its linear term coefficient will be 1.
To obtain the correct quadratic, we need to adjust the linear term so that it becomes 1x instead of -2x or 34x. To do this, we can divide each quadratic by the coefficient of its linear term.
For Joanie's quadratic:
Divide x^2 + 14x - 32 by -2 (the coefficient of the linear term):
(x^2 + 14x - 32) / -2 = -1/2x^2 - 7x + 16
For Kelvin's quadratic:
Divide x^2 + 34x - 72 by 34 (the coefficient of the linear term):
(x^2 + 34x - 72) / 34 = 1/34x^2 + 1x - 72/34 = 0.0294x^2 + x - 2.118
Now, both Joanie and Kelvin have factored the quadratics in monic form, where the coefficient of the quadratic term is 1. By comparing the quadratic terms, we can conclude that the quadratic the teacher wrote on the board is:
x^2 - 7x + 16