SOLUTION: I don't know if you have that textbook, but if so you may be able to help me. in section 8.1 example 4 and 8 there are examples where the number suddenly changes into i. i ll write
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Question 150883This question is from textbook intermediate algebra
: I don't know if you have that textbook, but if so you may be able to help me. in section 8.1 example 4 and 8 there are examples where the number suddenly changes into i. i ll write it down in case you don/t have the book.
the example 4:
(2x-5)2 = -16
2x-5 = +- square root -16
2x-5 = +- 4i
2x=5+-4i
x=5+-4i / 2
i have had the section about i in college but i don't understand the transition from root -16 to 4i. how does the i come into play?
8.1 example 8 is also sth with i, if you have the book it would be great if you could explain that to me as well.
anna This question is from textbook intermediate algebra
You can put this solution on YOUR website! (2x-5)^2 = -16
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i is the sqrt of -1.
That means -1 is a perfect square, just like 4 and 9 and 16 are perfect squares.
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Historically it was introduced into mathematics when someone needed a
solution to the following equation:
x^2+1 = 0
The answer had to be x^2 = -1
So x=i became the solution.
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Your Prolem:
(2x-5)^2 = -16
Take the square root of both sides to get:
2x-5 = +4i or 2x-5 = -4i
2x = 5 + 4i or 2x = 5 - 4i
x = (5+4i)/2 or x = (5-4i)/2
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I don't have your text so I can't talk about your other problem.
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Just remember that -1 is a perfect square; sqrt(-1) = i
Cheers,
Stan H.
You can put this solution on YOUR website! Dear Anna,
i is an imaginary number. you attain the square root of any negative number by taking the square root of the opposite and multiplying it by + or - i. So the square root of -16 is the square root of 16 (4 or -4) multiplied by i (4i or -4i). Since i^2=-1, +4i or -4i would be the sqaure root of -16, because:
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