3
votes
Accepted
Why won't factoring lead to the same result as expanding when simplifying $(x+y)^2 - y^2 - x^2 + 2xy$?
$$(x+y)^2 \color{green}{- y^2 - x^2} + 2xy=(x+y)^2 \color{red}{-(y+x)(y-x)} + 2xy \tag 1$$
It is a wrong claim,
$$RHS{=(x+y)^2 - (y+x)(y-x) + 2xy\\
=(x+y)^2 - (y^2-x^2) + 2xy\\
=(x+y)^2 - y^2+x^2 + ...
- 1,484
1
vote
Accepted
$\frac{1}{9-ab}+\frac{1}{9-bc}+\frac{1}{9-ca} \leq \frac{3}{8}.$
Let $$\sum_{cyc}(1-x)(k+x)\geq0$$ is true.
Thus, $$\sum_{cyc}(k+(1-k)ab-a^2b^2)\geq0$$ or after homogenization
$$k(a+b+c)^4-3(k-1)(ab+ac+bc)(a+b+c)^2-27(a^2b^2+a^2c^2+b^2c^2)\geq0,$$ which after ...
- 197k
1
vote
Accepted
How to solve quadratic equation with parameter for solutions in some range
I'm going to first point out an error in the method mentioned in question, and then show how to solve it using Vieta's formulas.
There is an error in $x_2=(p+3+p\color{red}+1)/4$.
It should be $x_2=(...
- 135k
1
vote
How to solve quadratic equation with parameter for solutions in some range
A vigilant student would have immediately noticed that $x_1=1$ is a root of the equation
$$2x^2-(p+3)x+p+1=0$$
and by polynomial division factorize the equation like
$$(x-1)(2x-p-1)=0.$$
Then the ...
- 8,694
1
vote
How to solve quadratic equation with parameter for solutions in some range
The question and the existing answer are too long, so I decided to write an answer after I read the first paragraph.
The discriminant $D$ of the equation equals $$(p+3)^2-4\cdot 2(p+1)=p^2-2p+1=(p-1)^...
- 85.8k
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