Amy, beth and Carla are solving a quadratic equation. Amy misreads the constant term and finds the roots 8 and 2. Beth misreads the coefficient of x and finds roots -9 and -1. Carla solves it. What are Carla's solutions?
Let the quadratic equation be
x²+bx+c=0
which factorizes to (if factorizable)
(x-p)(x-q)=0
This means that
pq=c, and p+q=-b
where p,q are the solutions to the equation.
If the constant term was misread, and assuming the coefficient of x was correctly read, then -b=(8+2), or b=-10.
If the coefficient of x was misread, and assuming the constant term was correctly read, then c=(-9*(-1))=9
The original equation is therefore:
x²-10x+9=0
which factorizes to
(x-9)(x-1)=0
Can you take it from here?
To find Carla's solutions, we need to correctly solve the quadratic equation using the given information from both Amy and Beth.
Let's start by setting up the quadratic equation in the standard form: ax^2 + bx + c = 0.
From Amy's misreading, we know that the roots are 8 and 2. Using the fact that the sum of the roots of a quadratic equation is equal to -b/a, we can deduce that -b/a = 8 + 2 = 10.
Similarly, from Beth's misreading, we know the roots are -9 and -1. So, -b/a = -9 + (-1) = -10.
Comparing the values of -b/a from the two scenarios, we can conclude that 10 = -10. However, this is not possible.
Therefore, there is a contradiction between the information given by Amy and Beth. It suggests that there is an error in either Amy's or Beth's reading or calculations.
Without additional information or clarification, we cannot determine Carla's solutions to the quadratic equation.