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What is the quadratic equation whose root is 1+3i?

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Final answer:

The quadratic equation with a root of 1 + 3i is x² - 2x + 10 = 0.

Step-by-step explanation:

This is a quadratic equation of the form ax² + bx + c = 0 where a, b, and c are constants. To find the quadratic equation with a root of 1 + 3i, we can use the fact that complex roots come in conjugate pairs. So the other root would be 1 - 3i. We can then use the quadratic formula to find the equation:

x = (-b ± √(b² - 4ac))/(2a)

Substituting the values a = 1, b = -2, and c = 10, we get:

x = (2 ± √((-2)² - 4(1)(10)))/(2(1))

Simplifying further, we have:

x = (2 ± √(4 - 40))/(2)

x = (2 ± √(-36))/(2)

x = (2 ± 6i)/(2)

Dividing both numerator and denominator by 2, we get:

x = 1 ± 3i

The quadratic equation with roots 1 + 3i and 1 - 3i is:

x² - 2x + 10 = 0

The quadratic equation whose root is 1+3i must also have 1-3i as its root since complex roots come in pairs. To find the equation, we use the fact that if a and b are roots of a quadratic then the equation is x² - (a + b)x + ab = 0. Substituting our roots in, the equation becomes:

x² - ((1 + 3i) + (1 - 3i))x + (1 + 3i)(1 - 3i) = 0 x² - (1 + 3i + 1 - 3i)x + (1 - (3i)²) = 0 x² - (2)x + (1 - (-9)) = 0 x² - 2x + 10 = 0

Therefore, the quadratic equation with the given root 1+3i is x² - 2x + 10 = 0.

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