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kvasek [131]
3 years ago
5

Solve each equation, if possible. Write irrational numbers in simplest radical form. Describe the strategy you used to get your

solution and tell why you chose that strategy.
a. x² +4 =0
b. x²^2 -6x +1=0
Mathematics
2 answers:
fomenos3 years ago
4 0
For the answer to the question above, I'll provide my solutions below.
x² + 4 = 0
<span>     - 4  - 4
</span>      x² = -4
       x = +2i

x² - 6x + 1 = 0
x = -(-6) +/- √((-6)² - 4(1)(1))
                       2(1)
x = 6 +/- √(36 - 4)
                 2
x = <span>6 +/- √(32)
</span>             2
x = <span>6 +/- 2√(2)
</span>             2
x = 3 + √(2)
x = 3 + √(2)    x = 3 - √(2)
I hope this helps.
schepotkina [342]3 years ago
4 0
Hello there.
<span>
Solve each equation, if possible. Write irrational numbers in simplest radical form. Describe the strategy you used to get your solution and tell why you chose that strategy.
</span>
x = 3 + √(2)    x = 3 - √(2)

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How many positive integers $n$ satisfy $127 \equiv 7 \pmod{n}$? $n=1$ is allowed.
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Naturally, any integer n larger than 127 will return 127\equiv127\mod n, and of course 127\equiv0\mod127, so we restrict the possible solutions to 1\le n.

Now,

127\equiv7\mod n

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In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,

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(2x^3)^2 x 6x^2 <br><br> (2m^0 x 5m)^2 x 5m^2
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\large\boxed{(2x^3)^2\times6x^2=24x^8}

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