Asked by Anissa
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?
Answers
Answered by
MathMate
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?
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?
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