hectictar

The solutions to  \(4x^2 + 3 = 3x - 9\)  can be written in the form  \(x = a \pm b i,\)  where  \(a\)  and  \(b\)  are real numbers. What is  \(a + b^2\) ?  Express your answer as a fraction.

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\(4x^2 + 3 = 3x - 9\)

                                        Subtract  3x  from both sides and add  9  to both sides of the equation.

\(4x^2-3x + 12 = 0\)

                                        Now we can use the quadratic formula to solve for  x

\(x\ = \ \frac{3\pm\sqrt{3^2-4(4)(12)}}{2(4)}\ =\ \frac{3\pm\sqrt{-183}}{8}\ =\ \frac{3\pm\sqrt{183}i}{8}\ =\ \frac38\pm\frac{\sqrt{183}}{8}i\)

And now we have the solutions in the form   \(x = a \pm b i\)   so we can see...

\(a+b^2\ =\ \Big(\frac38\Big)+\Big(\frac{\sqrt{183}}{8}\Big)^2\ =\ \frac38+\frac{183}{64}\ =\ \frac{24}{64}+\frac{183}{64}\ =\ \frac{207}{64}\)_

If we express  \(-2x^2 + 4x + 5\)  in the form  \(a(x - h)^2 + k\) ,  then what is  \(k\) ?

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=   -2x² + 4x + 5

                                     Factor  -2  out of the first two terms.

=   -2( x² - 2x ) + 5

                                            Add  1  and subtract  1  to complete the square inside the parenthesees.

=   -2( x² - 2x + 1 - 1 ) + 5

                                            Factor  x² - 2x + 1  as  (x - 1)²

=   -2( (x - 1)² - 1 ) + 5

                                     Distribute the  -2

=  -2(x - 1)² + 2 + 5

                                    Combine like terms.

=  -2(x - 1)² + 7

Now it is in the form  a(x - h)² + k  and we can see that  k = 7

Just one final thing in case it is unclear...

It is true that

\(\sum\limits_{k=1}^{1000} k(\lceil\log_{1.41421}k\rceil-\lfloor\log_{1.41421}k\rfloor)\ =\ 500499\)

It is also true that

\(\sum\limits_{k=1}^{1000}k(\lceil\log_{\sqrt2}k\rceil-\lfloor\log_{\sqrt2}k\rfloor)\ =\ 499477\)

The question asks for the second sum, not the first one.

\(y\ =\ \sqrt{3-\sqrt{x}}\\~\\ y\ =\ (3-x^\frac12)^\frac12\\~\\ y'\ =\ \frac{d}{dx}(3-x^\frac12)^\frac12\)

Using the power rule and chain rule,

\(y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ \frac{d}{dx}(3-x^\frac12)\ )\\~\\ y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ \frac{d}{dx}3-\frac{d}{dx}x^\frac12\ )\\~\\ y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ 0-\frac12x^{-\frac12}\ )\)

And we can rewrite the right side of the equation like this:

\( y'\ =\ \dfrac{-1}{4\,\cdot\,\sqrt{3-\sqrt{x}}\,\cdot\,\sqrt{x}} \)_

"for every number other than 1, the difference between the "ceiling" and the "floor" is always 1 (in this case)"

That is not true in the case where the base is  \(\sqrt2\)

See example:

\(\lceil\log_{\sqrt2}32\rceil-\lfloor\log_{\sqrt2}32\rfloor\ =\ 0\)

Okay I think I understand what is happening.

\(\sum\limits_{k=1}^{1000}\ k(\lceil\log_{1.41421}k\rceil-\lfloor\log_{1.41421}k\rfloor)\)     IS equal to  500,499

When  k = 1 ,   \(\lceil\log_{1.41421}k\rceil-\lfloor\log_{1.41421}k\rfloor\ =\ \lceil\log_{1.41421}1\rceil-\lfloor\log_{1.41421}1\rfloor\ =\ 0\)

But for every other value of  k , the difference is not zero.

For example:

\(\lceil\log_{1.41421}32\rceil-\lfloor\log_{1.41421}32\rfloor\ =\ 1\)

BUT

\(\lceil\log_{\sqrt2}32\rceil-\lfloor\log_{\sqrt2}32\rfloor\ =\ 0\)

So it makes perfect sense that

\(\sum\limits_{k=1}^{1000}\ k(\lceil\log_{1.41421}k\rceil-\lfloor\log_{1.41421}k\rfloor)\ =\ 500500-1\ =\ 500499\)

while

\(\sum\limits_{k=1}^{1000}\ k(\lceil\log_{\sqrt2}k\rceil-\lfloor\log_{\sqrt2}k\rfloor)\ =\ 500500-1023\ =\ 499477\)

Basically, if you estimate the base first you will not get the same answer.

I tried to state the base using the underscore but it didn't work, so I used this notation

log(b, a)

which WA will interpret as

\(\log_{b}a\)

When  k  is a power of  2 ,   \(\lceil \log_{\sqrt {2}}k\rceil - \lfloor \log_{\sqrt {2}}k \rfloor\ =\ 0\)

When  k  is not a power of 2 ,   \(\lceil \log_{\sqrt {2}}k\rceil - \lfloor \log_{\sqrt {2}}k \rfloor\ =\ 1\)

So  N  is the sum of all numbers which are not powers of  2  in the interval  [1, 1000]

\(N \quad=\quad \sum\limits_{k = 1}^{1000}k\quad-\quad \sum\limits_{n = 0}^{9}2^n\\~\\~\\ N \quad=\quad \sum\limits_{k = 1}^{1000}k\quad-\quad (1+2+4+8+16+32+64+128+256+512)\\~\\~\\ N \quad=\quad \sum\limits_{k = 1}^{1000}k\quad-\quad 1023\\~\\~\\ N \quad=\quad \Big(\frac{1000(1000+1)}{2}\Big)\quad-\quad1023\\~\\~\\ N \quad=\quad 500500\quad-\quad1023\\~\\~\\ N \quad=\quad 499477\)

Also since we know all the zeros are at   x = -2   and   x = 4 ,  we can say

y  =  a(x + 2)(x - 4)

Since the axis of symmetry on a parabola is always halfway between the roots, and since the maximum is 54....

vertex   =   ( \(\frac{4-2}{2}\), 54)   =   (1, 54)

Now that we know  (1, 54)  is a solution to  y = a(x + 2)(x - 4) , we can say

54  =  a(1 + 2)(1 - 4)

54  =  a( -9 )

-6  =  a

So the equation of the parabola is...

y  =  -6(x + 2)(x - 4)

y  =  -6( x² - 2x - 8 )

y  =  -6x² + 12x + 48

-6 + 12 + 48  =  54

Thanks.....there is still a LOT more I have to learn though....