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The values x = 2 ± i are the roots of the quadratic equation .

The values x = 2 ± i are the roots of the quadratic equation .-example-1
User JRoppert
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2 Answers

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Answer: The quadratic equation is
x^(2) -4x+5=0

Step-by-step explanation: First, let's find out the factors of the quadratic equation


(x-(2+i)(x-(2-i))=0

Simplifying further we get


[(x-2)^(2) -i^(2) ]=0\\{(x-2)^(2) +1]=0


(x-2)^(2) +1=0

The final Answer will be:


x^(2) -4x+5=0

User GTAVLover
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6 votes

Answer:

x^2 + 4x + 5 = 0

Explanation:

To determine the quadratic equation with roots x = 2 ± i, we can use the fact that complex roots always come in conjugate pairs. If one root is 2 + i, the other root will be its conjugate 2 - i.

To find the quadratic equation, we can use the fact that the sum and product of the roots of a quadratic equation are related to its coefficients.

Let's call the quadratic equation y = ax^2 + bx + c.

The sum of the roots is given by:

2 + i + 2 - i = 4

The product of the roots is given by:

(2 + i)(2 - i) = 4 - i^2 = 4 - (-1) = 5

Using these values, we can set up the equations:

Sum of roots:

b / a = 4

Product of roots:

c / a = 5

From the first equation, we can solve for b:

b = 4a

Substituting this into the second equation:

c / a = 5

We can choose any value for a, so let's set a = 1 to simplify the equation:

c = 5

Now we have the values of a, b, and c, which gives us the quadratic equation:

y = x^2 + 4x + 5 or x^2 + 4x + 5 = 0

User Narendra Motwani
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