Algebraic Techniques 01 - Perfect Square and Quadratic Equation

• For $1 \leq x < 2$, simplify the following expression.
• Solve the following equation.
• For real numbers $x$ and $y$, prove that
• Find real solutions of the following equation.
• Given the following inequality for all real numbers $x_1$, $x_2$, $\cdots$, $x_n$, find all positive integers $n$ where $n > 1$.
• Reference

For $1 \leq x < 2$, simplify the following expression.

$$\begin{aligned} \frac{1}{\sqrt{x + 2 \sqrt{x - 1}}} + \frac{1}{\sqrt{x - 2 \sqrt{x - 1}}} \end{aligned}$$

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[Solution] Check the denominators if they can changed to the perfect square form. $$\begin{aligned} x + 2 \sqrt{x - 1} &= x - 1 + 2 \sqrt{x - 1} + 1 = (\sqrt{x - 1} + 1)^2 \\ x - 2 \sqrt{x - 1} &= x - 1 - 2 \sqrt{x - 1} + 1 = (\sqrt{x - 1} - 1)^2 \end{aligned}$$

Since $x < 2$, $$\begin{aligned} \frac{1}{\sqrt{x + 2 \sqrt{x - 1}}} + \frac{1}{\sqrt{x - 2 \sqrt{x - 1}}} &= \frac{1}{|\sqrt{x - 1} + 1|} + \frac{1}{|\sqrt{x - 1} - 1|} \\ &= \frac{1}{\sqrt{x - 1} + 1} + \frac{-1}{\sqrt{x - 1} - 1} \\ &= \frac{-2}{(x - 1) - 1} \\ &= \frac{2}{2 - x} \end{aligned}$$

Solve the following equation.

$$\begin{aligned} x^4 - 97x^3 + 2012x^2 - 97x + 1 = 0 \end{aligned}$$

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[Solution] The big feature of this equation is that the coefficients are symmetric. As divided by $x^2$, $$\begin{aligned} x^2 - 97x + 2012 - \frac{97}{x} + \frac{1}{x^2} = 0 \end{aligned}$$

To make the perfect form, this can be rewritten to $$\begin{aligned} \left(x + \frac{1}{x}\right)^2 - 97\left(x + \frac{1}{x}\right) + 2010 = 0 \end{aligned}$$

Assuming that $y = x + 1/x$, $$\begin{aligned} y^2 - 97y + 2010 = 0 \\ (y - 30)(y - 67) = 0 \end{aligned}$$

So $y = x + 1/x = 30, 67$, $$\begin{aligned} x^2 - 30x + 1 &= 0, \quad x = 15 \pm \sqrt{224} \\ x^2 - 67x + 1 &= 0, \quad x = \frac{67 \pm \sqrt{4485}}{2} \end{aligned}$$

For real numbers $x$ and $y$, prove that

$$\begin{aligned} 3(x + y + 1)^2 + 1 \geq 3xy \end{aligned}$$

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[Solution] Aussuming that $x + y = a$ and $xy = b$, $$\begin{aligned} 3(a + 1)^2 + 1 \geq 3b \end{aligned}$$

Considering the quadratic equation $t^2 - at + b = 0$ whose two real roots are $x$ and $y$, its discriminant $\Delta = a^2 - 4b \geq 0$, which is $b \leq a^2 / 4$. Therefore, the problem can be proved by showing that the following inequality holds. $$\begin{aligned} 3(a + 1)^2 + 1 &\geq \frac{3a^2}{4} \\ 9a^2 + 24a + 16 \geq 0, &\quad (3a + 4)^2 \geq 0 \end{aligned}$$

Find real solutions of the following equation.

$$\begin{aligned} x^4 + 16x - 12 = 0 \end{aligned}$$

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[Solution] By observation, this equation has no cubic and quadratic terms. So this can be represented as $$\begin{aligned} x^4 + 16x - 12 &= \color{crimson}{(x^2 + a)^2 - (bx + c)^2} \\ &= x^4 + (2a - b^2)x^2 - 2bcx + a^2 - c^2 \end{aligned}$$

Comparing coefficients, $2a = b^2$, $16 = -2bc$, and $a^2 - c^2 = -12$. Since $a = b^2 / 2$ and $c = -8/b$, $$\begin{aligned} a^2 - c^2 = -12 \Longleftrightarrow \frac{b^2}{4} - \frac{64}{b^2} = -12 \end{aligned}$$

So $b$ can be set to $2$, and this makes that $a = 2$ and $c = -4$. $$\begin{aligned} x^4 + 16x - 12 &= (x^2 + 2)^2 - (2x - 4)^2 \\ &= (x^2 + 2x - 2)(x^2 - 2x + 6) = 0 \end{aligned}$$

Therefore, $x^2 + 2x - 2 = 0$ has solutions, $x = -1 \pm \sqrt{3}$, and $x^2 - 2x + 6 = (x - 1)^2 + 5 = 0$ has no solutions.

Given the following inequality for all real numbers $x_1$, $x_2$, $\cdots$, $x_n$, find all positive integers $n$ where $n > 1$.

$$\begin{aligned} x_1^2 + x_2^2 + \cdots + x_n^2 \geq x_n (x_1 + x_2 + \cdots + x_{n - 1}) \end{aligned}$$

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[Solution] This inequality can be rewritten as $$\begin{aligned} x_1^2 - x_1 x_n + x_2^2 - x_2 x_n + \cdots + x_{n - 1}^2 - x_{n - 1} x_n + x_n^2 \geq 0 \end{aligned}$$

By using some proper perfect square, $$\begin{aligned} &\left( x_1 - \frac{x_n}{2} \right)^2 + \cdots + \left( x_{n - 1} - \frac{x_n}{2} \right)^2 - \frac{n - 1}{4} x_n^2 + x_n^2 \geq 0 \\ &\Longleftrightarrow \left( x_1 - \frac{x_n}{2} \right)^2 + \cdots + \left( x_{n - 1} - \frac{x_n}{2} \right)^2 \geq \frac{n - 5}{4} x_n^2 \end{aligned}$$

If $n \leq 5$, the inequality holds. If $n > 5$, it does not hold when $x_1 = x_n / 2$, $\cdots$, $x_{n - 1} = x_n / 2$, $x_n = 1$. Therefore, the solutions are $n = 2, 3, 4, 5$.

Reference

[1] Titu Andreescu, 105 Algebra Problems.