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Why is cubic function one-one?

If $(x_1-x_2)(x_1^2+x_2^2+x_1x_2)=0$, then as you implied, either $x_1=x_2$ and we are done, or else $x_1\ne x_2$. Now it is enough to show that for every $x_1\in\mathbb{R}$ and for every $x_2\ne x_1$,...
J. Chapman's user avatar
2 votes

Why is cubic function one-one?

Yes, your proof is sufficient. If $x_1$ and $x_2$ satisfy $x_1^2+x_2^2+x_1x_2=0$, then you have shown that both: $x_1,x_2$ have opposite signs because $-x_1x_2=x_1^2+x_2^2>0$. $x_1,x_2$ have the ...
Dark Malthorp's user avatar
2 votes
Accepted

Used Chinese Remainder Theorem on this question, and I can't figure what went wrong

All your steps except the last one are correct. It is $$ 19056 \equiv 1731 \pmod{3465} $$ but not $$ 19056 \equiv 1236 \pmod{3465} $$
Reinhard Meier's user avatar
1 vote

How to Solve the Inequality $(-4-x) < \sqrt{x^2 + 4x - 5}$?

We first require $\sqrt{x^2 + 4x-5} \in\mathbb R$ or in other terms $x^2 + 4x-5 \geq 0$. That can be written as $(x+5)(x-1) \geq 0$ which gives that the solution set must be a subset of $$(-\infty, -5]...
Sahaj's user avatar
  • 3,076
2 votes

How to Solve the Inequality $(-4-x) < \sqrt{x^2 + 4x - 5}$?

As you noticed, $x^2+4x-5\geq 0\iff x\geq 1 \text{ or } x\leq -5 $ so that the square root (and thus the inequality) only makes sense in that case. The RHS is non-negative, so if the LHS is negative, ...
Julio Puerta's user avatar
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2 votes
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How to Solve the Inequality $(-4-x) < \sqrt{x^2 + 4x - 5}$?

HINT Let consider two cases: $-4-x <0$ (the inequlity is trivially true, when RHS exists) $-4-x \ge 0$ (we can square both sides)
user's user avatar
  • 156k
4 votes

Show $\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2}+\ldots+\sqrt{a_n^2+b_n^2} \geq \sqrt{\left(a_1+a_2+\ldots+a_n\right)^2+\left(b_1+b_2+\cdots+b_n\right)^2}$

First assume that $a_k, b_k \ge 0$. We need to prove that $$\left(\sum\limits_{k=1}^n \sqrt{a_k^2+b_k^2}\right)^2- \left(\sum\limits_{k=1}^n a_k\right)^2- \left(\sum\limits_{k=1}^n b_k \right)^2 \ge 0....
pie's user avatar
  • 6,173
4 votes

Show $\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2}+\ldots+\sqrt{a_n^2+b_n^2} \geq \sqrt{\left(a_1+a_2+\ldots+a_n\right)^2+\left(b_1+b_2+\cdots+b_n\right)^2}$

Let $\lVert \cdot \rVert$ be Euclidean norm on $\mathbb{R}^2$, and $v_i=(a_i,b_i)$. Then, the inequality can be written as $ \sum_{i=1}^n \lVert v_i \rVert \geqq \lVert \sum_{i=1}^n v_i \rVert $ ,...
HighwayStar's user avatar
1 vote

Numbers that are four times their decimal reverse

This is not a direct answer to the question, but a general way to see numbers defined by digit reversal and linear equality (it should work for any Presburger arithmetic but I don't have formalization)...
pcpthm's user avatar
  • 151
1 vote

counting possible coefficient of polynomial

Brute force search in this case is not difficult, because as you noted $r_1 + r_2 = m$, which is less than $20$. We only need to identify pairs of prime numbers (not necessarily distinct) whose sum is ...
Saeed's user avatar
  • 1,920
0 votes

The largest prime divisor of integer $x$ satisfying $\left[\frac x{1!}\right]+\left[\frac x{2!}\right]+\cdots+\left[\frac x{10!}\right]=1001$

$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left[\frac{{x}}{{n}!}\right]=\mathrm{1001} \\ $$ $${x}\left({e}−\mathrm{1}\right)\in\left[\mathrm{1001};\mathrm{1012}\right) \\ $$ $$\left[{x}...
yoga mulyadi's user avatar
2 votes

Solving $\frac{\sqrt{26 + \sqrt{x}} - \sqrt{26 - \sqrt{x}}}{\sqrt[3]{26 + \sqrt{x}} + \sqrt[3]{26 - \sqrt{x}}} = \frac{5\sqrt{2}}{4}$

HINT Let $u = \sqrt[6]{26 + \sqrt{x}}$ and $v = \sqrt[6]{26 - \sqrt{x}}$. Then the proposed equation is equivalent to \begin{align*} \frac{u^{3} - v^{3}}{u^{2} + v^{2}} = \frac{5\sqrt{2}}{4} \end{...
Átila Correia's user avatar
0 votes

Solving $\frac{\sqrt{26 + \sqrt{x}} - \sqrt{26 - \sqrt{x}}}{\sqrt[3]{26 + \sqrt{x}} + \sqrt[3]{26 - \sqrt{x}}} = \frac{5\sqrt{2}}{4}$

As several comments suggested, some context would be helpful. Are you looking for real or complex solutions? Anyway, as a first step toward a solution, try using the substitution $a=26-\sqrt{x}$ and $...
strugglingmathematician's user avatar
1 vote

Solve $\dfrac{1}{x}<4$

Consider the equation $\frac{1}{x} < 4$, then if $x >0$ we have $\frac{1}{x}\cdot x < 4\cdot x \implies 4x > 1 \implies x > \frac{1}{4}$ Now the other case is $x<0$, then we have $\...
Rishabh Kothary's user avatar
2 votes

Solve $\dfrac{1}{x}<4$

$\dfrac{1}{x}<4$ $\dfrac{1}{x}-4<0$ $\dfrac{1-4x}{x}<0$ then $0$ and $\frac{1}{4}$ are the critical points. $x<0$ or $x>\frac{1}{4}$
Lion Heart's user avatar
  • 7,213
2 votes

Solve $\dfrac{1}{x}<4$

If you want equivalent statements, you need to distinguish two cases: $$\frac{1}{x}<4\iff \left\{\frac{1}{x}<4\text{ and }x<0\right\}\text{ or }\left\{\frac{1}{x}<4\text{ and }x>0\right\...
Kolakoski54's user avatar
6 votes
Accepted

Maximize $f(x)=(1-x)^5(1+x)(1+2x)^2$

For $-1\le x \le-1/2$, applying the AM-GM inequality: $$\begin{align} f(x)&= 15^56^2\cdot\ \left( \frac{1-x}{15}\right)^5(1+x)\left(\frac{-1-2x}{6} \right)^2 \\ &\le 15^56^2\left(\frac{\frac{1-...
NN2's user avatar
  • 16.4k
1 vote
Accepted

Solve $\dfrac{x-1}{2}\geq \dfrac{2}{3} \geq 1-\dfrac{x}{6}$

I think what is perplexing you is you are expecting that because the question has been stated and a neat single line inequality, $a < b < c$, you expect the answer to do, $m < x < n$. so ...
fleablood's user avatar
  • 125k
1 vote

Maximize $f(x)=(1-x)^5(1+x)(1+2x)^2$

Using Log derivative : $$ d_{\log}f(x)=\dfrac{f'(x)}{f(x)}=5\dfrac{-1}{1-x}+\dfrac{1}{1+x}+2\dfrac{2}{1+2x}$$ $$f'(x)=f(x)\left(5\dfrac{-1}{1-x}+\dfrac{1}{1+x}+2\dfrac{2}{1+2x}\right)=f(x)\left(\dfrac{...
EDX's user avatar
  • 2,016
1 vote

Solve $\dfrac{x-1}{2}\geq \dfrac{2}{3} \geq 1-\dfrac{x}{6}$

You basically want to find an $x\in \mathbb{R}$ such that $3x-3\geq 4$ and $4\geq 6-x$ both hold. I plotted both inequalities for you : See $6-x\leq 4$ iff $x\geq 2$ but then $3x-3\leq4$ so this x won'...
muhammed gunes's user avatar
3 votes

Let $P \in \mathbb{R}[X]$ so that $P(0) = 0$ and $P(X+1) - P(X-1) = X^2 + X$. Find the value of $P(\frac{3}{2})$.

Alternativily we can directly calculate $P(2k)$, as you observed. As $P(2(k+1))-P(2k) = (2k+1)(2k+2)$ we can sum the left and right side and get a telescoping sum: $$ P(2k) = \sum_{m=0}^{k-1} P(2(m+1))...
zetko's user avatar
  • 91
2 votes
Accepted

Let $P \in \mathbb{R}[X]$ so that $P(0) = 0$ and $P(X+1) - P(X-1) = X^2 + X$. Find the value of $P(\frac{3}{2})$.

Hint: Assume $P(x)=ax^3+bx^2+cx$ and use $P(x+2)-P(x)=x^2+3x+2=a(x+2)^3+b(x+2)^2+c(x+2)-(ax^3+bx^2+cx)$ to get $a,b,c$. Edit: If $P(x)=dx^4+ax^3+bx^2+cx$, then the expanded form is $8dx^3+\ldots$, you'...
Bowei Tang's user avatar
  • 1,537
0 votes

Same day of the week birthdays

Rogerl has answered this but let's try a very simple approach with minimal maths. Your birthday is July 9. If your husband's was exactly a week later on July 16, would you be surprised that it was ...
badjohn's user avatar
  • 8,589
0 votes

Is $x^x = (x-1)^{x+1}$?

For the case where $n=1$, using @TravisWillse's approach (have a look here, we have $$x\sim e+\frac{3}{2}-\frac{7}{24 e}+\frac{1}{3 e^2}-\frac{911}{1920 e^3}+\frac{34}{45 e^4}-\frac{748075}{580608 e^5}...
Claude Leibovici's user avatar
1 vote

Let $ S=\{-1,0,\frac{1}{2},\sqrt{2},2 \}$. Solve for $(1)\ \dfrac{1}{x} \leq \dfrac{1}{2} \ \ \ \ (2)\ 10 \geq\dfrac{5}{2x}$.

It is permissible to directly substitute into the test. However, since the focus is on learning about this category of problems, a thorough understanding of fractional equations with unknowns in the ...
Frank's user avatar
  • 29
1 vote

$2^x + 3^x + 6^x = x^2$, no. Of solutions. (Without differentiation)

Obviously, $\color{blue}{x=-1}$ is a solution to the given equation $2^x+3^x+6^x=x^2$ as $\frac12+\frac13+\frac16=1$. Let's check out whether other real solutions also exist. First, note that the ...
Pustam Raut's user avatar
  • 2,147
1 vote

$2^x + 3^x + 6^x = x^2$, no. Of solutions. (Without differentiation)

$$2^x + 3^x + 6^x = x^2 \implies 6^x <x^2 \implies x<0$$ Hence assume $x>0$ but change equations to: $$2^{-x}+3^{-x} + 6^{-x} = x^2 \implies 3 \times 2^{-x} > x^2 > 2^{-x} \implies \...
Balaji sb's user avatar
  • 4,403
1 vote
Accepted

Simplify $\lfloor \frac{k}{10}\rfloor\cdot\lfloor \frac{k-10\cdot\lfloor \frac{k}{10}\rfloor}{5}\rfloor$

The 1st equation can be factorised into $$\begin{align*} \left\lfloor \frac{k}{10}\right\rfloor\cdot\left\lfloor \frac{k}{5}\right\rfloor-\left\lfloor \frac{k}{10}\right\rfloor^2+\left\lfloor \frac{k}{...
peterwhy's user avatar
  • 22.5k
13 votes

Numbers that are four times their decimal reverse

We can obtain the closed form expressions of those $n$-digit numbers as $\begin{align} 21 \underbrace{99\ldots9}_{n-4 \ \text{times}} 78 &= 22 \underbrace{00\ldots0}_{n-2 \ \text{times}} - 22 \\ &...
Euclid's user avatar
  • 706
2 votes
Accepted

Solve the equation for the variable $x$. $\sqrt{x}+a\sqrt[3]{x}+b\sqrt[6]{x}+ab=0$

$$\sqrt{x}+a\sqrt[3]{x}+b\sqrt[6]{x}+ab=0\\(a+\sqrt[6]{x})(b+\sqrt[3]{x} )=0$$ $a+\sqrt[6]{x}=0\implies\sqrt[6]{x}=-a<0$ since $a>0$. Therefore it is not a valid solution . or $b+\sqrt[x]{x}=0\...
Antony Theo.'s user avatar
  • 1,394
18 votes
Accepted

Finding the real roots of an octic polynomial (degree eight).

You're right: the polynomial initially resists simple attempts at finding its real roots through elementary methods like RRT (as a side note; you really only needed to try out the negative factors of ...
Sahaj's user avatar
  • 3,076
0 votes

Solve for $x$: $\ a^3x^3+b^3=0\ \ \ $ The constants $a,b$ represent positive real numbers.

$$b^3+a^3x^3=0\implies a^3x^3=-b^3\implies x^3=-\frac{b^3}{a^3}\\ $$ Taking the cube root of both sides gives $\sqrt[3]{-\frac{b^3}{a^3}}$ times the third roots of unity $$x=-\frac{b}{a}\lor x=\frac{\...
Antony Theo.'s user avatar
  • 1,394
2 votes
Accepted

Solve for $x$: $\ a^3x^3+b^3=0\ \ \ $ The constants $a,b$ represent positive real numbers.

You seem to be mistaking here: $$i^2=-1 \text{ (second power)}$$ $$(-1)^3 = -1 \text{ (third power)}$$
Dominique's user avatar
  • 2,434
0 votes

Solving $(x^2-1)^2-({x^*}^2-1)^2\geq{0}$

Find $t\in[-2,2]$ such that for all $x\in[-2,2],$ the inequality $$(x^2-1)^2-(t^2-1)^2\geq{0}$$ holds. Assume that $t\not\in\{-1,1\},\tag1$ and suppose that$\forall x{\in}[-2,2]\quad(x^2-1)^2-(t^2-1)^...
ryang's user avatar
  • 39.7k
0 votes

Proving $f(a)=1 \forall \; a≥ 1/8$ and its Relation to Cardano's Formula

Update: Refer here, to find the root of a depressed cubic equation $t^3 + pt + q = 0$, the method introduce two variables $u$ and $v$ such that $u+v=t$. Substitute this in the depressed cubic: $$u^{3}+...
eugene's user avatar
  • 33
4 votes
Accepted

Proving $f(a)=1 \forall \; a≥ 1/8$ and its Relation to Cardano's Formula

I don't know how to handle it by Cardano. Let $b= \sqrt{8a-1\over 3}$, then $$a = {3b^2+1\over 8}$$ So $$\left( a \pm \frac{\sqrt{\frac{8a - 1}{3}} \cdot \left( a + 1 \right)}{3} \right)^{\frac{1}{3}} ...
nonuser's user avatar
  • 90.5k
1 vote

Proving $f(a)=1 \forall \; a≥ 1/8$ and its Relation to Cardano's Formula

Let $x =\left( a + \frac{\sqrt{\frac{8a - 1}{3}} \cdot \left( a + 1 \right)}{3} \right)^{\frac{1}{3}}$ and let $y = \left( a - \frac{\sqrt{\frac{8a - 1}{3}} \cdot \left( a + 1 \right)}{3} \right)^{\...
Sahaj's user avatar
  • 3,076
1 vote

Proving $f(a)=1 \forall \; a≥ 1/8$ and its Relation to Cardano's Formula

$f(a)=1$ since if $-\frac q2=a$ and $t=1$ is a root of $t^3+pt+q=0$, i.e. $$\text{if}\quad q=-2a\quad\text{and}\quad p=2a-1,$$ then $$\frac{q^2}4+\frac{p^3}{27}=\frac{(a+1)^2(8a-1)}{27}.$$
Anne Bauval's user avatar
  • 38.8k
0 votes

Solving $(x^2-1)^2-({x^*}^2-1)^2\geq{0}$

$$|x^2-1|\ge|{x^*}^2-1|$$ for all x So what happens when $x=1$ $$ 0\ge|{x^*}^2-1| $$ This is only possible when ${x^*}^2=1$ The proper solution is to think of $|{x^*}^2-1|$ as separate constant say $y$...
RandomGuy's user avatar
  • 1,727
2 votes
Accepted

Showing $2(b-c)^2(d-a)^2 = (a-b)(a-c)(b-d)(c-d)$, given $(a+b)(c+d) = 2ab+2cd$

Consider $a,b,c$ as polynomial indeterminates, and define the rational fraction $d\in\Bbb Q(a,b,c)$ by $(a+b)(c+d)=2ab+2cd$, i.e. $$d:=\frac{2ab-ac-bc}{a+b-2c}.$$ Substituting $d$ in the identity we ...
Anne Bauval's user avatar
  • 38.8k
0 votes

Irrational equation problem

An elegant way of handling this uses the following identity. $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$ Now if you set $$a = \sqrt[3]{x+1} ; b = \sqrt[3]{x-1} ; c = -\sqrt[3]{5x}$$ the RHS is ...
Andreas's user avatar
  • 15.8k
0 votes
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Help finding the ellipse equation that passes through four points.

Hint: Substitute the point coordinates to get four equations (1) to (4). Now subtract (1) - (2) and (3) - (4) to get two equations which are linear in $h$ and $k$. Solve these, treating $a$ and $b$ as ...
mcd's user avatar
  • 3,526
0 votes

Help finding the ellipse equation that passes through four points.

(Not the exact way you wanted) I used python, like you did, but using minimization : ...
govindah's user avatar
  • 162
0 votes

Is $x^x = (x-1)^{x+1}$?

The ratio $(x-1)^{x+1}\over x^x$ is eay to estimate $\rho = (x-1).(1-{1\over x})^x=(x-1) e^{x \log(1-{1\over x})}$ Now $x \log(1-{1\over x})=-1-1/2{1\over x}+o({1\over x})$ So $\rho= (x-1)e^{-1} (1-1/...
Thomas's user avatar
  • 7,635
1 vote
Accepted

Question on exponential functions and drawing suitable lines on the graph

Well, raising to the power of $e$ first, we get $$ e^x = 4(1-x) $$ Multiply by $\frac 1 2$: $$ \frac 1 2 e^x = 2(1-x) $$ Add $3$: $$ \frac 1 2 e^x + 3 = 2(1-x) + 3 $$ Now simplify: $$ \frac 1 2 e^x + ...
PrincessEev's user avatar
  • 45.2k
2 votes

Is $x^x = (x-1)^{x+1}$?

This equation has many variables, so methods like Lagrange inversion or transforms would be hard to apply. However, series reversion works well. First, use @Claude Leibovici’s answer to notice that $x\...
Тyma Gaidash's user avatar
0 votes

Solve $16x^4=81$

Here's an interesting fact! For any $k\in\mathbb{Z},$ then $^{n}\sqrt{1}=e^{\frac{2\pi ik}{n}}.$ (These are called the $n^{\text{th}}$roots of unity.) Let $I_n$ denote the collection of all $n^{\text{...
JAG131's user avatar
  • 907
3 votes
Accepted

Solve $16x^4=81$

By the Fundamental theorem of algebra, if $p(x)$ is a polynomial with complex (possibly all real) coefficients of degree $n$, then equation $p(x) = 0$ has precisely $n$ solutions, counted with ...
Ennar's user avatar
  • 23.4k
2 votes

Solve $16x^4=81$

In order to avoid the case of only getting the two real solutions, we should keep in mind that any degree 4 polynomial with complex coefficients (i.e., on $\mathbb{C}[x]$) has 4 roots, counting the ...
南洋小學生's user avatar
0 votes

Solve $16x^4=81$

Your first approach is just fine, but remember: $$x^4=\frac{81}{16}\implies x=\frac{3}{2}\times\text{the $4^{th}$ roots of unity }$$ So $$x=-\frac{3}{2}\lor x=-\frac{3i}{2}\lor x=\frac{3i}{2}\lor x=\...
Antony Theo.'s user avatar
  • 1,394

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