I was trying to solve the following inequation: $$ x^2≤16 \Rightarrow x≤\sqrt{16} \Rightarrow x≤4 \text{ and } x≤-4 $$ But the solution given in my textbook is [-4,4], which is only possible when $x≥-4$ and $x≤4$. What have I done wrong?  (Sorry I could not format my question properly, I am new to this site.)
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Simplify: $x^2 &gt; 1$.  My solution: Taking square root on both sides:  $±x &gt; ±1$  So my results are:   $x &gt; 1$ $x &gt; -1$ $-x &gt; 1$ $\implies$ $(-1 &gt; x)$ $-x &gt; -1$ $\implies$ $(1 &gt; x) $   But I strongly feel this is wrong. What is wrong here?  A step-by-step explanation will help me.
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This came up while solving another ENT problem. I want to ask when is: $$2^n -7 \text{ where } n\geq 3$$ a perfect square? Specifically, I also wanted to know what would be the solutions when $n$ is odd? How should I solve this?  I can check that $n=3, 4, 7$ are solutions but cannot find more. As it is of the form $4k+1$, it doesn't help as well.
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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 ??? $
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Find all $x \in (-1, +\infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.  What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:   $(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y &gt; 1$, which can then be rewritten as:  $$ \left( \frac{y^2 + 1}{2y} \right)^{y - 2} - \left( \frac{y^2 - 1}{2y} \right)^{y - 2} = 1 $$ Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.  Do you have any suggestion on how I can continue?
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I want to know what will happen to the inequality in these cases  $a&gt;x&gt;b$ If $a$ and $b$ negative  When i square the both sides of it  Is it going to be  $b^2&gt;x^2&gt;a^2$  Ok but what would happen if  $a$ is a positive and $b$ is a negative number when i square ? EXAMPLE $2&gt;x&gt;-1$  another case  $2&gt;x&gt;-4$   i forgot  mentioning this case ?  If it is something like  $2&gt;x&gt;-2$
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Where is the mistake? $$-1=(-1)^{2/2}=\left(\left (-1\right)\displaystyle ^2\right)^{1/2}=1^{1/2}=\sqrt 1=1$$
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I can't seem to figure out the inequality option, I substituted n=2 for the answer bit is there any elegant way of getting it?
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Possible Duplicate:          I'm trying to understand the exact point of failure in the following reasoning:  \begin{equation*} 1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{\sqrt{-1}^2\sqrt{-1}^2} = \sqrt{(\sqrt{-1}\sqrt{-1})^2} = \sqrt{-1}\sqrt{-1} = \sqrt{-1}^2 = -1. \end{equation*}  I've been previously told that the problem is due to square root not being a function in C; which I found totally unhelpful. Could someone please explain the problem here in simpler terms.  Edit:  Thank you all for your comments in trying to help me understand this. I finally do. Following is the explanation on my problems in understanding this, in case it'll be of any help to anyone else.  My problem was really due to using an incorrect definition of i: $i = \sqrt{-1}$. While the correct definition would be: $i^2 = -1$.  My incorrect definition led me reasoning, such as (which superficially seemed to give expected results. I see now that this is incorrect, too):  \begin{equation*} \sqrt{-9} = \sqrt{9 * (-1)} = \sqrt{\sqrt{9}^2 \sqrt{-1}^2} = \sqrt{(\sqrt{9} \sqrt{-1})^2} = \sqrt{9} \sqrt{-1} = 3i. \end{equation*}  Instead, had I used the correct definition of i:  \begin{equation*} {(xi)}^2 = x^2i^2 = -x^2 = -9, \\ x^2 = 9, \\ x = +- 3. \end{equation*}  Now, analyzing the equations in the original problem, I can see at least the following two errors:  1) In the third =, I'm relying on $-1 = {\sqrt{-1}}^2$, while I should be relying on: $-1 = (+-\sqrt{-1})^2$ which would of course give two different branches. Hmm.. on the second reading, this isn't really a problem, as even with the two separate branches, both of them will lead to the result in the next step.  2) In the fifth =, I'm relying on $\sqrt{i^4} = i^2$, which would be correct, if i was a non-negative number in R. But as i is the imaginary unit and in C: $\sqrt{i^4} = \sqrt{i} = +-(1/\sqrt{2})(1 + i)$.
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$|x|=\sqrt{x^2}$ as  shows. But, as $(x^2)^\frac12=x$, I can't understand where am I wrong interpreting Square-root.
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If a>0 and b>0,  $ \lim_{x\rightarrow +\infty } \sqrt[x]{a^x+b^x} = ? $  What I was trying to do: Suppose a>b. Then, for sufficiently large values of x, $ a^x &gt;&gt; b^x $; so $\sqrt[x]{a^x+b^x} \rightarrow \sqrt[x]{a^x} \rightarrow a $ when $x \rightarrow +\infty$.   Is that idea correct? How can I formalize it?
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I am curious about whether $|x|^p$ with $0&lt;p&lt;1$ satisfy $|x+y|^p\leq|x|^p+|y|^p$ for $x,y\in\mathbb{R}$.   So far my trials show that this seems to be right... So can anybody confirm whether this is right or wrong and give a proof or counterexample?
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I'm trying to solve this equation:  $$2-x=-\sqrt{x}$$ Multiply by $(-1)$: $$\sqrt{x}=x-2$$ power of $2$: $$x=\left(x-2\right)^2$$ then: $$x^2-5x+4=0$$ and that means: $$x=1, x=4$$    But $x=1$ is not a correct solution to the original equation.  Why have I got it? I've never got a wrong solution to an equation before. What is so special here?
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this is something that came up when working with one of my students today and it has been bothering me since.  It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.  Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?  With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers?
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Could someone please explain how to solve this : $x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$ - not the answer only, but a step-by-step solution.  I tried to solve it, with the help of khanacademy, but still I have no idea how to correctly solve it.  Thank you so much in advance!
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Some one please help me with this question. I am getting -10.5 as the wrong final answer.I found the formula in the following slide.Since the result was coming out in complex number i got its modulus.That is how i ended up with a 10.5 (2 * 5.25).
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Hey so while doing exercises this popped up. My idea is to put everything under $\sqrt{x+1}$ and then multiply it by $\sqrt{x+1}$. and then square both sides to get that $x \leq 0$. But this is a very ugly method. The one in the answer is using the fact that $4(x^2+x) \leq (2x+1)^2$ but how in the world am i supposed to come up with that... This is actually the result i come up with at the end. It seems as if they did my method and then reverse it. Any other good ideas to solve it? Not graphically. Is it actually viable to calculate the function of area between the functions and maximize the function? Seems ugly here but in general?
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How $\sqrt{x}$ can be a function when $\sqrt{4}$ is equal to $-2$ and $2$?