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Valentin [98]
2 years ago
15

What is f(-3) for the function f(x) = -5x - 7? ​

Mathematics
1 answer:
sergij07 [2.7K]2 years ago
7 0
F(x) = -5(-3) - 7
15 - 7
f(-3) = 8
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What are the types of roots of the equation below?<br> - 81=0
Tju [1.3M]

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,

  • the equation x⁴ - 81 = 0

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

  1. (x² -  9) = (x² - 3²) = (x - 3)(x + 3) [using algebraic identity, (a + b)(a - b) = a² - b²]
  2. 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

Learn more about roots of equation here:

brainly.com/question/26926523

#SPJ9

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