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zaharov [31]
3 years ago
6

What is the value of n in the equation 0=18+128n^2?

Mathematics
1 answer:
lbvjy [14]3 years ago
7 0

Answer:

\large\boxed{\text{no soltion in the set of real numbers}}\\\boxed{n=-\dfrac{3}{8}i\ \vee\ n=\dfrac{3}{8}i\ \text{in the set of complex numbers}}

Step-by-step explanation:

18+128n^2=0\qquad\text{subtract 18 from both sides}\\\\128n^2=-18\qquad\text{divide both sides by 128}\\\\n^2=-\dfrac{18}{128}\\\\n^2=-\dfrac{18:2}{128:2}\\\\n^2=-\dfrac{9}{64}

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Let x=the first number
Then x+7=second number
And 4(x+7)+2=third number (2 more than 4 times the second)
Now we are told that:
x+(x+7)+(4(x+7)+2)=85 get rid of parentheses
x+x+7+4x+28+2=85 simplify
6x+37=85 subtract 37 from each side
6x+37-37=85-37
6x=48
x=8---first number
x+7=8+7=15 second number
4(x+7)+2=4*15+2=62 third number
CK
8+15+62=85
85=85
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