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.
What are the types of roots of the equation below?
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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