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We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

By Cauchy-Schwarz, the minimum vaue is n/(n + 1).

By AM-GM, the maximum value is 5/8.

The inverse of 2 modulo 4 is 17, so n = 15*17 = 255 = 20 (mod 47).

Here's a link to your answer

https://web2.0calc.com/questions/sequence_38

Here's a answer to your question:

https://web2.0calc.com/questions/quadratics_17

True.  I should have noted that the factors could be -1 and -47 as well as 1 and 47.

Hi Alan

I already had  

(24,23)

(-24,-23)

you added

(24,-23)

I guess  (-24,23)  can also be added.

So now we have 4 points  

Like so (first considering the numerators and denominators separately):

I get two lattice points:

Let 7x^2 + 5x = h + 3x^2 - 11x.  What value of will h give us exactly one solution for x?

Hello Guest!

$4x^2+16x-h=0\\ x^2+4x-\frac{h}{4}=0\\ x=-2\pm\sqrt{4+\frac{h}{4}}$

$4+\dfrac{h}{4}=0\\ 4=-\dfrac{h}{4}\\ \color{blue}h=-16$

h = - 16 will  give us exactly one solution for x.

  !

What is the coefficient of x^3 when 24x^4 + 6x^3 + 4x^2 + 7x - 5 is multiplied by 6x^3 + 3x^2 - 3x + 4 and the like terms are combined?

Hello Guest!

$(24x^4 + 6x^3 + 4x^2 + 7x - 5) ( 6x^3 + 3x^2 - 3x + 4)$

$=144x^7+108x^6-30x^5+132x^4+3x^3-20x^2+43x-20$

$=x(x(x(x(x(x(144x+108)-30)+132)+3)-20)+43)-20$

  !

The possible outcomes are:

a  1  and a  5

a  2  and a  4

a  3  and a  3

Of these 3 possible outcomes, there is 1 where the number three appears at least once.

And so the probability that the number three appeared at least once  =  1/3

Mmm

1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484,529,576

1+47=47

4+47+51

9+47=56

16+47=63

25+47=72

36+47=83

49+47=96

64+47=111

81+47=128

100+47=147

121+47=168

144+47=191

169+47=216

196+47=243

225+47=272

256+47=303

289+47=336

324+47=371

361+47=408

400+47=447

441+47=488

484+47=531

529+47=576    I already had that one.

I do not think there are any others.

Assuming that x is an integer, values of x are 6 through 18

The cases are {4,0,0}, {3,1,0}, {2,1,1}, {2,2,0}

Counting the number of ways in each case gives us an answer of 1 + 4 + 6 + 6 = 17.

((39-3)/3 +1)! = 13! Ways

7+12>x

x+12>7

x+7>12

=>...

x>5

x<19

(5,19)

z^2 = 7

z^2 - 7 = 0

sum: 0

4x^2+16x-h=0

4(x^2 + 4x - x/4) = 0

x^2 + 15x/4=0

x = - 15/4

That was my originaly answer and it's wrong apparently

Here's a link to your answer:

https://web2.0calc.com/questions/help-please_82361

Using the triangle inequality, x must be bigger than 7 + 12, so x > 19

Find the constant coefficient when the polynomial 3(x - 4) + 2(x^2 - x + 7) - 5(x - 1) + 8(x^2 - x + 3) - 4(x - 5) is simplified.

Look at each bracket separately, find the constants, then add them together.

3(x-4)  constant will be 3 * -4=-12

+ 2(x^2 - x + 7) constant will be  +2 * +7 = +14

- 5(x - 1)  constant will be  -5 * -1 = +5

etc

mmm

$x^2-y^2=47\\ x^2=47+y^2\\ 47+y^2=x^2$

so 47+ some square number = another square number

here are the smallest square numbers.  I will stop when the difference between the biggest and the one just below it is more than 47

1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484,529,576

576-529=47    so I found one (actually two).  24^2-23^2=47

(24,23)

(-24,-23)

No bigger squares can work

Maybe there are some others in there, you can look, but I doubt it.