cryptoaops

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Given \(x^2+2y^2+2xy+4y+7\)

\(x^2+2y^2+2xy+4y+7\)

\(x^2+y^2+2xy+y^2+4y+7\)

\(x^2+y^2+2xy+y^2+4y+4+3\)

\((x+y)^2+(y+2)^2+3\)

Minimum of square of any number is \(0\). So, minima will be at \(x=2,y=-2\). Therefore, \((2+(-2))^2+(-2+2)^2+3=3\). Therefore, the least value of the expression \(x^2+2y^2+2xy+4y+7\) is \(\boxed3\).

Good Job, CalTheGreat (flec)!

How many 4-digit positive integers exist that satisfy the following conditions:

(A) Each of the first two digits must be 1, 4, or 5, and

(B) the last two digits cannot be the same digit, and

(C) each of the last two digits must be 5, 7, or 8?

(A)
Each of the first two digits must be 1, 4, or 5 ?

\(\begin{array}{|lllc|c|} \hline & x&x &xx \\ \hline &1 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & 3\times 100 \\\\ &4 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & +3\times 100 \\\\ &5 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & +3\times 100 \\\\ \hline & & & & = 3\times 3\times 100 = {\color{red}900} \\ \hline \end{array}\)

There are 900 4-digit positive integers that satisfy the condition.

(B)
The last two digits cannot be the same digit ?

\(\begin{array}{|lcc|c|} \hline & xx &xx \\ \hline &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 00 & 90 \\\\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 11 & 90 \\\\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 22 & 90 \\\\ & \cdots \\ \\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 99 & 90 \\\\ \hline & & & = 90\times 10 = 900 \\ \hline \end{array}\)

1000 until 9999: There are 9000  four-digit integers.

9000 - 900 = 8100

There are 8100 4-digit positive integers that satisfy the condition.

(C)
Each of the last two digits must be 5, 7, or 8 ?

\(\begin{array}{|lclc|c|} \hline & x&x &xx \\ \hline &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 5& 100\times 3 \\\\ &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 7& +100\times 3 \\\\ &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 8& +100\times 3 \\\\ \hline & & & & = 3\times 100\times 3 = {\color{red}900} \\ \hline \end{array}\)

There are 900 4-digit positive integers that satisfy the condition.

This must be an Alcumus question. Please do not cheat on your Alcumus. Otherwise, you could possibly be reported to the AoPS Student Services.

Answer:​ \(\boxed{x = -5, \frac12}\)

Step-by-step explanation: The given equation is \(2x^2 + 9x - 5 = 0\)

We have to find the values of \(x\).

We know that if in a quadratic equation \(ax^2+ bx + c = 0, b^2- 4ac > 0\) then there will be two real roots of the equation.

In the given equation \(b^2- 4ac = (9)^2-4(2)(-5) = 81 + 40\)

\(= 212 > 0\)

So there will be two real roots.

Now we will factorize the given equation.

\(2x^2 + 9x - 5 = 0\)

\(2x^2 + 10x - x - 5 = 0\)

\(2x(x + 5) - 1(x + 5) = 0\)

\((2x - 1)(x + 5) = 0\)

So two roots will be \(2x - 1 = 0\)

\(x = \frac12\)

and \(x + 5 = 0\)

\(x = -5\)

Therefore, \(x = -5, \frac12\),  are the solutions.

\(y=x^2\\ y=\pm \sqrt{x}\)

The coordinate plane is divided into five planes by the graphs \(y=x^2\) and \(y=\pm\sqrt{x}\).

Solution: \(\mathrm{Range\:of\:}\frac{5x^2+20x+23}{x^2+8x+5}:\quad\begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le\frac{-\sqrt{421}+16}{11}\quad\mathrm{or}\quad\:f\left(x\right)\ge\frac{\sqrt{421}+16}{11}\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty\:,\:\frac{-\sqrt{421}+16}{11}]\cup\:[\frac{\sqrt{421}+16}{11},\:\infty\:)\end{bmatrix}\)

Steps: Rewrite \(\frac{5x^2+20x+23}{x^2+8x+5}\)as \(\frac{5x^2+20x+23}{x^2+8x+5}=y\)

\(=\frac{5x^2+20x+23}{x^2+8x+5}=y\)

Multiply both sides by \(x^2+8x+5\).

\(=\frac{5x^2+20x+23}{x^2+8x+5}\left(x^2+8x+5\right)=y\left(x^2+8x+5\right)\)

Simplify \(\frac{5x^2+20x+23}{x^2+8x+5}\left(x^2+8x+5\right):5x^2+20x+23\)

\(=5x^2+20x+23=y\left(x^2+8x+5\right)\)

The range is the set of y for which the discriminant is greater or equal to zero

Discriminant \(5x^2+20x+23=y\left(x^2+8x+5\right):\quad44y^2-128y-60\)

\(44y^2-128y-60\ge\:0\quad:\quad y\le\frac{-\sqrt{421}+16}{11}\quad\) or \(\quad\:y\ge\frac{\sqrt{421}+16}{11}\)

Check if the range interval endpoints are included

\(y=\frac{-\sqrt{421}+16}{11}\quad:\quad\)Included

\(y=\frac{\sqrt{421}+16}{11}\quad:\quad\)Included

Therefore, the range is \(f\left(x\right)\le\frac{-\sqrt{421}+16}{11}\quad\) or \(\quad\:f\left(x\right)\ge\frac{\sqrt{421}+16}{11}\).

Solution #1: If the numbers are called \(x\) and \(y\), then \(y = x - 9\), and so the sum of squares

\(x^2+ y^2 = x^2 + (x-9)^2 = 153\)

\(x^2 + (x^2 - 18x + 81) = 153\)

\(2x^2 - 18x - 72 = 0\)

\(x^2 - 9x - 36 = 0\)

This is factorable. However, if you use the quadratic formula rather than just factoring, the formula shows that either \(x=12\) and \(y=3\), or \(x=-3\) and \(y=-12\). Either way, \(x\cdot y = \boxed{36}\).

Solution #2: \(x-y = 9…(1). x^2+y^2 = 153…(2). (2)-(1)^2\rightarrow2xy = 153–81 =72\text{,} \text{ ie., } xy = \boxed{36}.\)

Both ways, you get the same answer of \(\boxed{36}\).