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Solve $x^2+6x-15120=0$

I'm hesitant to answer the question you asked because 1) we have no reason to assume it can be factored and if it can't that's a waste of a lot of time and 2) you should use the quadratic formula. ...
fleablood's user avatar
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6 votes
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Solve $x^2+6x-15120=0$

You noticed that the two roots are of the form: $-u,v$ such that $$u-v=6\text{ and }uv=2^4 \cdot 3^3 \cdot 5 \cdot 7.$$ You conjectured that $u,v$ are integers. At least one of them is then even (...
Anne Bauval's user avatar
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Solve $x^2+6x-15120=0$

By the Rational Root Theorem, any rational roots must be factors of 15120, i.e., in the set $\pm \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 27, 28, 30, 35, 36, 40, 42, 45, 48, ...
Dan's user avatar
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-1 votes

Solve $x^2+6x-15120=0$

To answer your question, it's necessary that the factors $(x+a),$ $(x+b)$ must be derived via $a,b\in\mathbb{C}$ such that $a$ and $b$ both divide $15120.$ (So you can derive the $a,b$ component of ...
JAG131's user avatar
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2 votes

Finding all real values of $k$ so that the eqn. $(x^2-x+1)^2- 2kx (x^2+x+1) + (x^2+x+1)^2=0...$

Another way: Check that $f(x)=0$ can be simplified to $$x^4-kx^3+(3-k)x^2-kx+1=0 \implies (x^2+1/x^2)-k(x+1/x)+(3-k)=0.$$ Let $x+1/x=z$, then we get $h(z)=z^2-kz+1-k=0$ for all real non-zero values of ...
Z Ahmed's user avatar
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2 votes

Finding all real values of $k$ so that the eqn. $(x^2-x+1)^2- 2kx (x^2+x+1) + (x^2+x+1)^2=0...$

A number $x$ is a root iff $x\ne0$ and $k=f(x)$, where $$f(x)=\frac{(x^2-x+1)^2+(x^2+x+1)^2}{2x(x^2+x+1)}.$$ Let $$t:=x+\frac1x.$$ Then, $|t|\ge2$, and each $t$ corresponds to two distinct values of $...
Anne Bauval's user avatar
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5 votes
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Finding all real values of $k$ so that the eqn. $(x^2-x+1)^2- 2kx (x^2+x+1) + (x^2+x+1)^2=0...$

A pretty generalizable approach is to use the discriminant or resultant. Here $$ R := \operatorname{Res}(f, f') = -128(k^2 + 4k - 4)^2 (3k - 5)(k + 5). $$ Now we know that $f$ has a double root if and ...
Sera Gunn's user avatar
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2 votes
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Let $\alpha, \beta$ be the roots of $x^2-a_1x+1=0$, and consider the sequence of numbers $a_r(r\ge0)$ with $a_0=1$ and $a^2_{r+1}=1+a_r.a_{r+2}$

The claim holds for $n=0$ and $n=1$, since $a_1=\alpha+\beta$, by well-known relations between roots and coefficients. Then we prove it by induction. Having $a_n=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-...
Massimiliano Foschi's user avatar
1 vote
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Quadratic where roots and coefficients together form Arithmetic Progression

Your equations are correct, and there are no more if we try to extend to the complex domain. Given the case $b=0$, we must have $a=-2d,m=-d,b=0,n=d,c=2d;$ and then the second Vieta formula $mn=c/a$ ...
Oscar Lanzi's user avatar
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0 votes

Find the canonical form of each equation and classify the quadratic curve

Write the linear terms in terms of $\sqrt 5 x+\frac{2y}{\sqrt 5}$ and $\frac{6y}{\sqrt 5}$. Something like $$a\left(\sqrt 5 x+\frac{2y}{\sqrt 5}\right)+b\frac{6y}{\sqrt 5}=-32x-56 y$$ Then complete ...
Andrei's user avatar
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5 votes

Quadratic with integer roots

My initial answer overlooked an analytical shortcut that may be used to moderately streamline the demonstration. I have added an Addendum, at the end of my answer, to provide the shortcut. $$x^2 -...
user2661923's user avatar
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6 votes
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Quadratic with integer roots

The following is taken from Equation and integers on AoPS. Let $p, q$ be positive integers such that $f(x) = x^2 - pq x + (p+q)$ has integral roots. Then $f(0) = p + q > 0$, so that necessarily $f(...
Martin R's user avatar
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Why do we need this inequality?

So you were right to say k>−1/4, you say your confused with how e^-x >0, This is simply due to the fact that e^x for any real value of x, is positive. Since the equations are based on e^-x, ...
Marx Carton's user avatar
2 votes
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Quadratic equation with a relation between its coefficients

The second one goes like this. Given: $$4a+4b+7c=0$$ $$\implies 4a+4b+4c=-3c$$ $$\implies a+b+c=-3c/4$$ $F(0)=c$ and $f(1)=a+b+c$. So, $f(0)\cdot f(1) = c \cdot(a+b+c) = -3c^2/4$. Which is always ...
Yashwanth's user avatar
0 votes

Quadratic solution from quartic

Observe $ \begin{align} G = \pm \Big( \omega C - \frac {1}{\omega L} \Big). \quad & \Longleftrightarrow \quad \omega C - \frac {1}{\omega L} = \pm G. \\ & \Longleftrightarrow \quad \omega - \...
Simon's user avatar
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-1 votes

Quadratic solution from quartic

$$\frac{Gw}{C}=\pm(w^2-\frac{1}{LC})\implies w^2-\frac{1}{LC}=\pm \frac{Gw}{C} \leftrightarrow w^2 \pm\frac{Gw}{C}-\frac{1}{LC}=0$$
Bowei Tang's user avatar
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1 vote

Is it possible for a quadratic function to not have $y$-intercept?

In order for a quadratic function $f(x)=ax^2+bx+c$ to not have a $y$-intercept, it's necessary that $f(0)$ must not exist (DNE). Seeing that $f(0)=a(0)^2+b(0)+c=c$, it follows that $f(0)$ necessarily ...
JAG131's user avatar
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1 vote
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Solve the given quadrilateral

As $DAC$ is isoceles, $\angle DAC=55°$. So by cosines law, $$x^2+AC^2-2xAC\cos 55°=x^2\implies AC=2x\cos 55°$$ $\angle CAB=5°$ again by law of cosines, $$AC^2+100-20AC\cos 5°=x^2\\\implies 4x^2\cos^...
Gwen's user avatar
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6 votes

why is it called quadratic formula?

A square has four sides, but its area is the side length raised to the second power. This is why raising to the second power ("squaring") is intimately linked to the Latin word for 4. And I ...
Arthur's user avatar
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4 votes

why is it called quadratic formula?

Historically, quadratic equations were viewed as equalities about areas of squares and rectangles. They were even treated differently depending on what we now call the signs of the coefficients. For ...
Ian's user avatar
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2 votes

Discriminant formula - do the coefficients include their signs?

Just rewrite everything in general form $$ ax^2+bx+c=0 $$ You can use round brackets if helpful $$(a)x^2+(b)x+(c)=0$$ Eg1 $$ 2x^2 - 2x = 0 $$ Rewrite $$0 = 2x^2 - 2x$$ $$= (2)x^2 - 2x$$ $$= (2)x^2 + -...
BCLC's user avatar
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1 vote
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If $f:\mathbb{R}-\{-1,K\}\rightarrow\mathbb{R}-\{\alpha,\beta\}$ is a function given by $f(x)=\dfrac{(2x-1)(2x^2-4 px+p^3)}{(x+1)(x^2-p^2x+p^2)}$

(1) I think your approach is correct. (In the two subcases, you might want to write that $f(x)$ is a bijective function.) (2) I think the given solution is wrong since they forgot to consider the ...
mathlove's user avatar
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