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.)
it is wrong!
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.)
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.)
I got the same question
lol I know im saying this 5 years later.
To find the roots of the quadratic that the teacher wrote on the board, let's first denote the given quadratic as x^2 + bx + c, where b is the coefficient of the linear term and c is the constant term.
Joanie's quadratic has roots -16 and 2, so she has factored it as (x + 16)(x - 2) = 0. If we multiply this out, we get x^2 + 14x - 32 = 0. Therefore, the quadratic that Joanie copied has a linear term coefficient of 14 and a constant term of -32.
Kelvin's quadratic has roots -36 and 2, so he has factored it as (x + 36)(x - 2) = 0. If we multiply this out, we get x^2 + 34x - 72 = 0. Therefore, the quadratic that Kelvin copied has a linear term coefficient of 34 and a constant term of -72.
Now, let's find the roots of the quadratic that the teacher wrote on the board, denoted as x^2 + bx + c.
To find the coefficient of the linear term b, we can add Joanie's and Kelvin's linear term coefficients and subtract the constant term c. So, b = 14 + 34 - (-32) = 14 + 34 + 32 = 80.
To find the constant term c, we can multiply Joanie's and Kelvin's constant term coefficients. So, c = -32 * (-72) = 2304.
Therefore, the quadratic that the teacher wrote on the board is x^2 + 80x + 2304 = 0.
To find the roots of this quadratic, we can factor it or use the quadratic formula. However, since the quadratic is a perfect square trinomial, we can rewrite it as (x + 40)^2 = 0. Setting this expression equal to zero, we have (x + 40)^2 = 0. Taking the square root of both sides, we get x + 40 = 0. Solving for x, we have x = -40.
So, the roots of the quadratic that the teacher wrote on the board are -40 and -40 (since it is a perfect square quadratic). Therefore, the roots are -40, -40.
Joanie's equation is
(x+16)(x-2) = 0
x^2 + 14x - 32 = 0 , in hers the -32 is wrong, the 14x is correct
Kelvin's equation is
(x+36)(x-2) = 0
x^2 + 34x - 64 = 0
for his, the 34x is wrong , but the -64 is right
the teacher's equation is
x^2 + 14x - 64 = 0