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Give an example of a quadratic function that has two real solutions with a multiplicity of 2, and solve for them algebraically.

A. f(x) = (x - 2)^2, solutions: x = 2 (multiplicity 2)
B. f(x) = (x + 3)^2, solutions: x = -3 (multiplicity 2)
C. f(x) = (x^2 - 4), solutions: x = ±2 (multiplicity 2)
D. f(x) = (x^2 + 1), solutions: x = ±1i (multiplicity 2)

User Nembleton
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1 Answer

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Final answer:

An example of a quadratic function with two real solutions with a multiplicity of 2 is f(x) = (x - 2)², which yields the solution x = 2. Since this is a perfect square, the graph touches the x-axis at x = 2 but does not cross it, confirming the multiplicity.

Step-by-step explanation:

To find a quadratic function that has two real solutions, each with a multiplicity of 2, we need to look for a function that will touch the x-axis at one point and not cross it. This occurs when the function is a perfect square trinomial. An example would be:

A. f(x) = (x - 2)2, solutions: x = 2 (multiplicity 2)

To solve for the solutions algebraically, we set the function equal to zero:

0 = (x - 2)2

We then take the square root of both sides:

±√0 = x - 2

Since the square root of 0 is 0, this simplifies to:

0 = x - 2

Adding 2 to both sides gives us the solution:

x = 2

Since this is the only solution we obtained, and we arrived at it by solving a perfect square, this solution indeed has a multiplicity of 2, which means the graph of the function touches the x-axis at this point but does not cross it.

User Maninalift
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