Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0. This can be obtained by finding root of the equation using algebraic identity.
<h3>What are the types of roots of the equation below?</h3>
Here in the question it is given that,
By using algebraic identity, (a + b)(a - b) = a² - b², we get,
⇒ x⁴ - 81 = 0
⇒ (x² + 9)(x² - 9) = 0
⇒ (x² + 9)(x² - 9) = 0
- (x² - 9) = (x² - 3²) = (x - 3)(x + 3) [using algebraic identity, (a + b)(a - b) = a² - b²]
- x² + 9 = 0 ⇒ x² = -9 ⇒ x = √-9 ⇒ x= √-1√9 ⇒x = ± 3i
⇒ (x² + 9) = (x - 3i)(x + 3i)
Now the equation becomes,
[(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0
Therefore x + 3, x - 3, x + 3i and x - 3i are the roots of the equation
To check whether the roots are correct multiply the roots with each other,
⇒ [(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0
⇒ [x² - 3x + 3x - 9][x² - 3xi + 3xi - 9i²] = 0
⇒ (x² +0x - 9)(x² +0xi - 9(- 1)) = 0
⇒ (x² - 9)(x² + 9) = 0
⇒ x⁴ - 9x² + 9x² - 81 = 0
⇒ x⁴ - 81 = 0
Hence Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: What are the types of roots of the equation below?
x⁴ - 81 = 0
A) Four Complex
B) Two Complex and Two Real
C) Four Real
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