Questions about quadratic functions and equations, second degree polynomials usually in the forms $y=ax^2+bx+c$, $y=a(x-b)^2+c$ or $y=a(x+b)(x+c)$.

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In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use ...
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Prove: For odd integers $a$ and $b$, the equation $x^2 + 2 a x + 2 b = 0$ has no integer or rational roots.

If $a$ and $b$ are odd integers, prove that the equation $$x^2 + 2ax + 2b = 0$$ has no integer or rational roots.
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Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$.

Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$. I made quite a few attempts but could not ...
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Where did $-4x$ come from?

I'm going over my quadratic equations for the ACT and I came across this quadratic: $$(x – 2)^2 – 12$$ My teacher said we could have factored it out into this: $$x^2 – 4x – 8$$ But I just don't ...
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Determine all the values of the parameter $a$ for which the inequality $3-|x-a|>x^2$ is satisfied by at least one negative $x$.

I wanted to know, how can I determine all the values of the parameter $a$ for which the inequality $3 - |x-a| > x^2$ is satisfied by at least one negative $x$. I tried for $x<a, |x-a|=-(x-a)$ ...
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Two circles intersection

Could you tell what are all the four points in following? Two circles intersect at two points maximum when we want to draw intersecting circles. But there we are solving quadratic equations, what is ...
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I have a inequality, I don't know where to start

Show that for $x,y,z > 0$ the inequality is true: $\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+x+y+z \geq \frac{(x+y)^2}{y+z}+\frac{(y+z)^2}{z+x}+\frac{(z+x)^2}{x+y}$ I have tried Holder, but i had ...
Intuition for why equations of the form $k^x=x^c$ are not solvable trivially?
It recently occurred to me that I did not know how to solve equations of the form $k^x=x^c$ for any two constants $k$ and $c$. After much pain in algebraically manipulating the equation (using ...
find $f(\frac{1}{2014})+f(\frac{2}{2014})+…+f(\frac{2013}{2014})$ of $f(x)=\frac{2}{2+4^x}$
$f(x)=\frac{2}{2+4^x}$ find $f(\frac{1}{2014})+f(\frac{2}{2014})+.....+f(\frac{2013}{2014})$ Please guide me through it, the only step I know is probably to eliminate the denominator ps. Not a ...