14
votes
Why do we sometimes lose solutions when solving equations?
One frequent cause of losing solutions is dividing by something that could be $0$. For example, you get an equation of the form
$$ a b = b c $$
and you jump to the conclusion $a = c$ without taking ...
- 441k
12
votes
Why do we sometimes lose solutions when solving equations?
If that is the case why do we sometimes lose solutions when solving equations?
We don't. If we take an equation and perform valid operations on it, we will never lose any solutions. If we look back ...
- 9,849
8
votes
Accepted
Erroneous application of AM-GM inequality.
Equality for this expression (and hence maximum for the geometric mean) is attained when $a=b=c$.
This is where the error lies. For example, pick any three positive numbers $a'\ge b'\ge c'$ that are ...
- 2,456
7
votes
Express $(989)\cdot(1001)\cdot(1007) + 320$ as product of primes.
A possible exploratory approach which is lucky to work:
The average of $989$ and $1001$ and $1007$ is $999$ so the product is slightly less than $999^3$
If $n=999$ then the product is $(n-10)(n+2)(n+...
- 154k
7
votes
Accepted
If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$
Some thoughts.
By Holder inequality, we have
\begin{align*}
&\left(\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a^2 + 4bc}} \right)^2\cdot \sum_{\mathrm{cyc}} (a^2 + 4bc)(4b + 4c - bc + 4ab + 4ac)^3 \\
\...
- 36.8k
7
votes
Accepted
Summation inside tan(x)
In fact, rewriting $\tan(a_k)$ under the form :
$$\tan(a_k)=\frac{\tfrac{k}{n}-\tfrac{k-1}{n}}{1+\tfrac{k}{n}\tfrac{k-1}{n}}$$
we recognize in the RHS the formula of
$$\tan(a-b)=\frac{\tan(a)-\tan(b)}{...
- 79k
7
votes
Why do we sometimes lose solutions when solving equations?
TL;DR (short response)
$$ ``\text{operation"} \neq ``\text{function"}. $$
But if we remove references to functions from the question, the question is a good one.
Discussion
The problem with the ...
- 95.9k
6
votes
Accepted
Find minimum of odd quadratic expression
Hint: Put $\alpha=\cos^{2}(\theta)$. The function becomes $A\cos (2\theta)-B\sin(2 \theta)$. Use C-S inequality to show that the minimum is $-\sqrt {A^{2}+B^{2}}$.
- 31.5k
6
votes
Polynomial problem with vieta formulas
Assuming $a_0=1$
Let $\alpha_1,\alpha_2,...,\alpha_{20}$ be the roots of $P(x)=0$
$\Rightarrow\alpha_1+\alpha_2+...+\alpha_{20}=0$
Also, $\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+...\frac{1}{\alpha_{20}}=...
- 8,652
6
votes
Erroneous application of AM-GM inequality.
First,
$$ (x-3)(2y+1)(3z+5) = 6 (x-3)(y+\frac{1}{2})(z+\frac{5}{3}). $$
Then, by AM-GM,
$$ (x-3)(y+\frac{1}{2})(z+\frac{3}{5}) \leqslant (\frac{1}{3}((x-3)+(y+\frac{1}{2})+(z+\frac{5}{3})))^3 = (\frac{...
- 83
6
votes
Why do we sometimes lose solutions when solving equations?
We lose solutions for the same reason that we lose keys, or lose our glasses: it is because we are not careful. If we are careful when solving equations, we will not lose solutions.
- 177k
5
votes
If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$
Some thoughts.
By AM-GM, it suffices to prove that
$$\frac{2}{\frac{a^2 + 4bc}{2 + bc/2} + (2 + bc/2)} + \frac{2}{\frac{b^2 + 4ca}{2 + ca/2} + (2 + ca/2)} + \frac{2}{\frac{c^2 + 4ab}{2 + ab/2} + (2 + ...
- 36.8k
5
votes
Knowing $x^2-x-1$ is a factor of $p(x)=ax^5 + bx^4 + 1$ , find a,b.
A simple long division method yields
$p(x)=ax^5+bx^4+1=(x^2-x-1)((ax^3)+(a+b)x^2+(2a+b)x+(3a+2b))+\ (5a+3b)x+(3a+2b+1)$
as we know $(x^2-x-1)$ is factor of $p(x)$, The remainder $(5a+3b)x+(3a+2b+1)$ ...
- 909
4
votes
$ab+bc+ca+abc=4$, prove $\sum\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8}$
Sketch of a proof.
We use the so-called isolated fudging.
It suffices to prove that
$$\frac{2+\sqrt{ab}}{\sqrt{ab}+c} + \frac{a^2 + b^2}{16abc} \ge \frac{39}{8}\cdot \frac{6a + 6b + abc}{12(a + b + c) ...
- 36.8k
4
votes
Is this proof that $0^0=1$ valid?
Let me answer your title question: Is this proof valid?
No, it is not.
On line 1 of your proof, you are making an implicit assumption, namely that $x$ is an element of the domain of the function $x^x$....
- 117k
4
votes
How to tell the question between subtraction and negative numbrers in algebra
I’m not sure if I’m understanding correctly, but I think you’re saying that you don’t know whether an expression like “$5-3$” should be read as “five minus three” or as a “five times negative 3”. If ...
- 2,610
4
votes
Accepted
Can polynomial $p(x)$ have integer roots?
You were almost there!
Continuing from where you left off, from the equation
$$
q(x_0) = -\frac{1}{(x_0 - x_1)(x_0 - x_2)(x_0 - x_3)}
$$
it follows that
$$
-\frac{1}{(x_0 - x_1)(x_0 - x_2)(x_0 - x_3)}...
- 58k
4
votes
Determine equation of circle tangent to both the $x$-axis at $(4, 0)$ and the line $y=x+1$ (and perhaps modify question to be a better exercise)
The intersection point of the tangents is $(-1,0)$ From this we know the length of tangent is 5 units.
Now we can join this point to the centre $(4,r)$ and form a right triangle. As one of the ...
- 136
4
votes
Accepted
How to evaluate $\lim_{x \to 0} \frac{\tan{x} - \sin{x}}{x(1 - \cos{x})}$?
$\frac{\tan{x} - \sin{x}}{x(1 - \cos{x})}=\frac {\sin x} x\frac {\frac 1 {\cos x}-1} {1-\cos x}=\frac {\sin x} x \frac 1 {\cos x}$ so the limit is $1$.
- 31.5k
4
votes
Accepted
Inequality $\frac{a}{\sqrt{b}} + \frac{b}{\sqrt{a}} \ge \sqrt[4]{8}$ on the unit circle
Using the substitutions $a = x^2$ and $b = y^2$ where $x, y \in (0, 1)$, we have $x^4 + y^4 = 1$, so
$$
\frac{a}{\sqrt{b}} + \frac{b}{\sqrt{a}}
= \frac{x^2}{y} + \frac{y^2}{x}
= \frac{x^3 + y^3}{x y}.
...
- 1,887
4
votes
Accepted
Is $e^{\ln(-7) }= -7$?
There are two good answers I can see, I will explain both.
If we define the functions $\ln x$ and $e^x$ for real inputs and outputs, there is a problem with $\ln x$ when $x$ is negative. This can be ...
- 5,328
4
votes
Accepted
Find a sum that consists of roots of polynomial and it's derivative polynomial
It is correct that
$$
\frac{f'(x)}{f(x)} = \sum_{i=1}^{n} \frac{1}{x - \alpha_i}
$$
for all $x \notin \{ \alpha_1, \ldots, \alpha_n \} $.
Since $f$ has $n$ distinct roots, $f$ and $f'$ have no common ...
- 109k
4
votes
$a*b=a \iff a=0$ or $b=1$
You should base your proof on:
$ab = a \iff$
$ab - a = 0 \iff$
$a \cdot (b-1) = 0$
The multiplication of two integers is zero is (at least) one of the integers is zero.
- 1,599
4
votes
Accepted
When and why is the sum of infinite zeroes exactly zero?
Your example can be written as
$$
L = \lim_{n\to\infty} \sum_{r=1}^\infty f_r(n)
$$
with $f_r(n) = \frac{r}{n^2+r^2} \,\mathbf 1_{n\geq r}$, and your question is a subcase if the question: when can we ...
- 11.9k
4
votes
Equation of plane $\mathbf{x} = (1, 0, 1) + s(1, 3, -1) + t(2, 2, 1)$
First: Since $(1,3,-1)$ and $(2,2,1)$ are vectors parallel to the plane. We can use their cross product to find the components of the vector $\vec{n}=(a,b,c)$, which is perpendicular to the plane as ...
4
votes
Seeking help determining the limit of a convergent sequence
The expression $ x _ n ^ 3 - ( n ^ 2 + 3 n ) x _ n - 3 n ^ 2 $ is written as a polynomial in $ x _ n $, but since $ n $ (not $ x _ n $) is diverging to infinity, let's rewrite it as a polynomial in $ ...
- 3,538
4
votes
If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$
Disclaimer: Not a full solution
If one of $a, b, c=0$:
WLOG let $a=0$. Since $ab+bc+ca>0$ we know that $b, c\neq0$.
$$\begin{align}
& \frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ca}}+\frac{1}{\...
- 2,940
4
votes
If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$
Remark 1: This is my third proof (the idea) which is better than my two old proofs.
Remark 2: Some years ago, I used a similar idea for the problem:
Let $a, b, c > 0$ with $a + b + c = 3$. Prove ...
- 36.8k
4
votes
Accepted
Find the value of $\log_{5}(0.0016)$
Your working is fine and correct.
Here is a shorter working:
$$y=\log_5(0.2)^4=4\log_5(5^{-1})=-4$$
- 148k
4
votes
Accepted
Knowing $x^2-x-1$ is a factor of $p(x)=ax^5 + bx^4 + 1$ , find a,b.
Your method is on-target. You just need to reduce the quintic until both the equations are of same degree to compare coefficients. I will show an example with highest degree of $2$ (reduction to ...
- 2,352
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