Question #82350

2 Answers

Answer:

Neither of you is correct. The real zeros of this polynomial are 1 and 4.

Explanation:

The easiest way to see this is by using the quadratic formula.
If you have a polynomial in the form of #ax + by + c#, then you can find both the real zeros like this:
1) # (-b-sqrt(b^2-4ac))/(2a) #

2) # (-b+sqrt(b^2-4ac))/(2a) #

In this case #a = -1#, #b = 5# and #c = -4#.
Plugging this in the formula you get:

1) #(-5-sqrt(5^2-4*(-1)*(-4))) / (2(-1)) = 4 #

2) #(-5+sqrt(5^2-4*(-1)*(-4))) / (2(-1)) = 1 #

So remember this formula:

# (-b+-sqrt(b^2-4ac))/(2a) #

Jun 19, 2015

Answer:

There is no answer, where there is no question.

Explanation:

#-x^2+5x-4# is an expression, it is not a question.

If the question is: "Evaluate #-x^2+5x-4# when #x=1#, then your teacher is correct.

If the question is: "For what value of #x# does the expression evaluate to #-868#, then you are correct

If the question is something else, (Like where is the vertex of #y=-x^2+5x-4#) then it is possible that neither of you is correct.