Your numbers are 2, 3, 4.
The three consecutive numbers are x, x+1, x+2
The square of the first equals the third so x^2=x+2
Solve this quadratic as x^2-x-2=0
Factor into (x+1)(x-2)=0
So your answers are x=-1 and x=2
Since the question says they’re positive integers, use x=2
So the numbers are 2, 3, 4
The Quadratic Function has the domain as the set of all real numbers.
For the range, start from minimum value to maximum value.
But because the parabola is downward as a < 0. Thus, there are no minimum value but the maximum value instead.
Therefore the range is y <= -4
X = 3/2
<em><u>I hope this could help!</u></em>
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Answer:
Step-by-step explanation:
The Fundamental Theorem of Algebra states that the number of complex roots a polyomial has is equal to its highest exponent. This is a squared polynomial; second degree; quadratic. When it is factored, no matter what types of numbers you get as the solution, you will ALWAYS have 2 of them. When this quadratic is factored, we get that x = 3 and x = 3. That means that this is a quadratic that touches the x-axis at (3, 0). It doesn't go through, it only touches. We do have 2 roots, but since they're the same, we say we have a multiplicity 2 of that root. The closest you'll come to that in your choices is A. Apparently your text refers to multiplicity 2 as a double root.
