The results I get not the same as in the book, basically I need to get only this and the $x$'s of course:  $$y = \frac{\sqrt{2}}{2}$$ So this is what I did: $$\sqrt{i} = x+yi$$ $$i = x^2 +2xyi -y^2$$  $$\begin{cases} x^2-y^2 = 0 \\  2xy = 1 \end{cases}$$  $$xy=0.5$$  $$x=\frac{0.5}{y}$$  $$(1)\space\space\space\space(\frac{0.5}{y})^2-y^2=0$$ $$0.25-y^4=0$$ $$y^4=0.25$$ $$y=\pm \sqrt[4]{0.25}$$ $$y=\pm \frac{\sqrt{2}}{2}$$  What is WRONG?
41442
If $i=\sqrt{-1}$, is $\large\sqrt{i}$ imaginary?     Is it used or considered often in mathematics? How is it notated?
47452
Frist I solved this equation $\displaystyle \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{xy}}$ by multiplying $xy \left(\frac{xy}{x}+\frac{xy}{y}\right)=y+x$ The answer is correct.  Next equation was:    $$\frac{\frac{1}{x^2}-\frac{1}{y^2}}{\frac{1}{x^2}+\frac{1}{y^2}}$$  I made the bottom multiplied and got $x^2 y^2.$ So I made the up also multiply and got     $$\frac{\frac{y^2}{x^2}-\frac{x^2}{y^2}}{\frac{1}{x^2y^2}}.$$   Now I'm stuck.  So now I doubt I did the first one the right way. I might have gotten the answer right but not the correct way to get it. Im sorry if Im confusing. Can someone help me with the two were I go wrong?
71323
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive   $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$  $$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$  Is the proof valid?
21179
It's not too difficult to figure out that $x$ and $y$ can both be 1, and also $x$ can be 2 and $y$ can be 4 (and vice versa). But I can't rule out if there are other solutions. Does it have anything to do with inverse functions? Is there a way to see the solutions by graphing, or algebraically?
27302
I am so confused,  how to get a solution of following differential equation:  $$(y')^2 -xy' + y = 0$$  My Thought:  Try solving $y'$ to get $y' = \dfrac{ x \pm \sqrt{x^2 - 4y}}{2}$ and now its unsolvable. Kindly help, appreciated.
56629
I'm wondering if there is a mathematical way of finding the roots of $y=x^3+x^2-6x-7$?  Supposedly, the roots are $2\cos\left(\frac {4\pi}{19}\right)+2\cos\left(\frac {6\pi}{19}\right)+2\cos\left(\frac {10\pi}{19}\right)$, $2\cos\left(\frac {2\pi}{19}\right)+2\cos\left(\frac {14\pi}{19}\right)+2\cos\left(\frac {16\pi}{19}\right)$ and $2\cos\left(\frac {8\pi}{19}\right)+2\cos\left(\frac {12\pi}{19}\right)+2\cos\left(\frac {18\pi}{19}\right)$.  Anything helps. I don't think substituting $x$ with something like $t+t^{-1}$ will work.
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Solve in $\mathbb{R}$ $\begin{cases} &amp;x+y+\frac{x^2}{y^2}=7 ~\cdots (\text{I})\\ &amp;\frac{(x-y)x^2}{y^2}=12~~~~~ \cdots (\text{II})\\ \end{cases}$ A friend's attempt: $ I.(x-1)= x^2 - 7x + 12= y^2 - 7y  \rightarrow x = y+ k \\  (y+k)^2-7(y+k)+12= y^2-7y \\ y = \dfrac{-k^2+7k-12}{2k} \rightarrow x = \dfrac{k^2 + 7k - 12}{2k} \\ \dfrac{k \cdot (k^2+7k-12)^2}{(-k^2+7k-12)^2 } = 12\\  k^5+2k^4+193k^3-1044k^2+2160k-1728=0 ??? $
83316
I am trying to solve a 2 part question, the first part is to expand $\sqrt{1-x}$ up to and including the term $x^2$ which I did.  This gives me $1-(0.5)x-(0.125)x^2 + ...$  However, the 2nd part of the question says that by taking $x=(1/64)$ in the expansion of part (i) -> which refers to the one above, deduce that $\sqrt 7 = 10837/4096$. How do i do this part?  EDIT (After Hint Given By SimpliFire):  $$ \sqrt{1-\frac1{64}}=\frac{\sqrt{63}}8=\frac38\sqrt7 $$ $$ =\frac83\sqrt{1-\frac1{64}}\;=\sqrt7 $$ $$ =\frac83\left(1-\frac12\left(\frac1{64}\right)-\frac18\left(\frac1{64}\right)^2\right)\; $$ $$ =\frac83\left(1-\frac1{128}-\frac1{32768}\right)\; $$ $$ =\frac83\left(\frac{32768}{32768}-\frac{256}{32768}-\frac1{32768}\right)\; $$ $$ =\frac83\left(\frac{32511}{32768}\right)\; $$ $$ =\frac{260088}{98304} $$ $$ \frac{10837}{4096}\;\approx\sqrt7 $$  I'm not sure if I can present it like this, if it is okay i will post it as my answer :X
89309
$Let$ $a,b,c&gt;0$  $$\begin{cases}\log_a(b^x)=2\\\log_b(c^x)=2\\\log_c(a^x)=5\end{cases}$$  $$x=?$$  So my attempt is just to use the logarithmic definition:  $$\log_a(b^x)=2\iff b^x=a^2$$  By similar logic,  $$a^x=c^5$$ $$c^x=b^2$$  So, if add everything together, we should get:  $$a^x+b^x+c^x=a^2+b^2+c^5$$  Looks to me like x is equal 2 different numbers at the same time which is strange, what am I doing wrong here?  I'm going to be a maths student in the upcoming year, this is taken from the Tel-Aviv university preparation material - shouldn't be too complex.
24351
Actually that is, I would like to use a complex exponent to turn a number x into an -x just by applying some complex exponent to x. I think I would have to use some ln, π and so one, but I am actually stack and cannot find the correct path to solving it.  For example, let's say for 2 we choose the real part to be equal 1, what would be the solution for y in this special case.  $$ 2^{1 + iy}=-2 $$  What would be the solution for the imaginary part of z if the real part must be 0.5?  $$ x^{0.5 + iy} = -x $$  $$ x^{0.5} * x^{iy} = \sqrt{x} * -\sqrt{x} = -x $$  What is y in this case?  Is there maybe a general solution for  $$ x^{z}=-x $$
82158
Proof  of identity $\sqrt {xy} = \sqrt x \sqrt y$ for $x,y \in \mathbb R^+$  I've been looking at the stated identity, which makes sense in $\mathbb R^+$ but fails in $\mathbb R$, since $\sqrt {-1 \cdot -1} \neq \sqrt {-1} \sqrt {-1}$.  How does one prove this identity ?  Suppose we have $x,y \in \mathbb R^+$ then $\sqrt{xy}^2 = xy = (\sqrt x \sqrt y)^2$.  Idea 1: The square-root function is bijective (monotonic increasing) and has inverse $X^2$. This in turn implies $\sqrt {xy} = \sqrt x \sqrt y$  Idea 2: The equation $X^2 = xy$ has at most $2$ solutions in $\mathbb C$, which implies $\sqrt x \sqrt y$ must be equal to $\pm \sqrt {xy}$.  Are these ideas rigorous enough ? Is there some simpler way of proving this ?
99252
I have equations \begin{align} 7z'(x) &amp;= x^2 + y(x)^2 + z(x)^2  \\ y''(x) &amp;= -(y'(x) + 7y(x))\sin(z(x)) \end{align} where $$(y(0),y'(0),z(0))=(1.5,-2.6,0.5)$$  The task is to solve this problem using a computer. Can someone help with explaining what goes in $u'(x)$?  Thanks.  I am guessing $u(x) = (y'',z')$.
114141
Q1 - Why is $\sqrt{-x}*\sqrt{-x}=-x?$  Q2 - I was thinking it would be:  $\sqrt{-x}*\sqrt{-x}=\sqrt{-x*-x}=\sqrt{x^2}$ but apparently not (why not?)  Q3 - What are the formal algebra rules to use? Can I calculate this without using i such as in: $\sqrt{-x}*\sqrt{-x}=i\sqrt{x}*i\sqrt{x}=-\sqrt{x^2}=-x$.
116259
$|x|=\sqrt{x^2}$ as  shows. But, as $(x^2)^\frac12=x$, I can't understand where am I wrong interpreting Square-root.
66332
Here is what I did, tell me whether I did correct or not: \begin{align*} y &amp;= \cos A + \cos B + \cos C\\ y &amp;= \cos A + 2\cos\left(\frac{B+C}2\right)\cos\left(\frac {B-C}2\right)\\ y &amp;= \cos A + 2\sin\left(\frac A2\right)\cos\left(\frac {BC}2\right) &amp;&amp; \text{since $A+B+C = \pi$} \end{align*}  Now for maximum value of $y$ if we put $\cos\left(\frac {B-C}2\right) = 1$ then \begin{align*} y &amp;\le \cos A + 2\sin\left(\frac A2\right)\\ y &amp;\le 1-2\sin^2\left(\frac A2\right) + 2\sin\left(\frac A2\right) \end{align*}  By completing the square we get  $$y \le \frac 32 - 2\left(\sin\frac A2 - \frac 12\right)^2$$  $y_{\max} = \frac 32$ at $\sin\frac A2 = \frac 12$ and $y_{\min} &gt; 1$ at $\sin \frac A2&gt;0$ because it is a ratio of two sides of a triangle.  Is this solution correct? If there is a better solution then please post it here. Help!
10869
No matter how I do it, I always end up with $x = 0, 90, 270$ and $360$. All of those except $270$ is right, but I can't quite figure out how to get the $270$ degrees out of the answer. I've tried using trig identities, I've tried squaring both sides, but I always end up with $$2\sin x\cos x$$ which then leads me to $x = 0, 90, 270, 360$. But $$\sin (270) + \cos (270) = -1$$ so I'm doing something wrong.
3981
As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.  But then I remember that $\sqrt(x)$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$.   So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.  But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.  Where does the absolute value in this derivation of the calculation come from?  I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there.