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? (Write your answers in increasing order, separated by commas.)

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To find the roots of the quadratic that the teacher wrote on the board, we need to consider the information given by Joanie and Kelvin.

Joanie correctly factors her quadratic and determines that it has roots -16 and 2. This means that her quadratic can be written in factored form as (x + 16)(x - 2) = 0.

Kelvin correctly factors his quadratic and determines that it has roots -36 and 2. This means that his quadratic can be written in factored form as (x + 36)(x - 2) = 0.

Now, we can compare the two factored forms we obtained. By equating the corresponding factors, we can determine the actual quadratic equation.

Setting the factors equal to each other, we have:
(x + 16)(x - 2) = (x + 36)(x - 2)

Expanding both sides, we get:
x^2 + 16x - 2x - 32 = x^2 + 36x - 2x - 72

Combining like terms, we have:
x^2 + 14x - 32 = x^2 + 34x - 72

Subtracting x^2 from both sides and simplifying, we get:
14x - 32 = 34x - 72

Subtracting 34x from both sides and simplifying further, we get:
-20x = -40

Dividing both sides by -20, we get:
x = 2

Therefore, the correct coefficient for the linear term is 2, and the quadratic equation is:
x^2 + 2x - 32 = 0

Now we can find the roots of the quadratic by solving this equation. To do this, we can use the quadratic formula.

The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the roots are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 1, b = 2, and c = -32. Plugging these values into the quadratic formula, we have:
x = (-2 ± √(2^2 - 4(1)(-32))) / (2(1))
x = (-2 ± √(4 + 128)) / 2
x = (-2 ± √132) / 2
x = (-2 ± 2√33) / 2

Simplifying further, we can divide both the numerator and denominator by 2:
x = -1 ± √33

Therefore, the roots of the quadratic that the teacher wrote on the board are -1 - √33 and -1 + √33, in increasing order.

So, the roots of the quadratic are:
-1 - √33, -1 + √33