hairyberry

C)

We know a-b = 1, and a+b = 1 or a+b = -1.

Consider two cases:

Case 1:

\(\begin{cases} a-b = 1\\ a+b=1 \end{cases}\)

\(2a = 2, a=1\)

\(\begin{cases} a=1 \\ b=0 \end{cases}\)

Case 2:

\(\begin{cases} a-b=1 \\ a+b = -1 \end{cases}\)

\(2a=0, a=0\)

\(\begin{cases} a=0 \\ b=-1 \end{cases}\)

Our two solutions are \(a, b = (1, 0)\) and \(a, b = (0, -1)\).

B)

\({(a+b)}^2={(a-b)}^{2}+4ab\)

From previously, ab=0

\((a+b)^2={1}^{2}\)

\(a+b=\pm1\).

A)

\({(a-b)}^{3}= {a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3} = 1\)

\(1-3{a}^{2}b+3a{b}^{2}=1\)

\(3ab(b-a)=0\)

\(-3ab=0\)

\(ab=0\)

If a number is a square of a binomial, it's discriminant is zero.

\(\Delta = {17}^2-4p=0\).

\(4p = {17}^2\)
\(p=\frac{{17}^{2}}{4}=\frac{289}{4}\).

The probability that Alice wins a game is the probability she will win on her first turn + the probability that she will win on her second turn + the probability she will win on her turn turn, and so on...

Alice has a \(\frac{1}{2}\) probability of winning on their first turn, finishing the game.

Since whenever someone wins, the game is finished Bob has a \(\frac{1}{2}\cdot\frac{1}{2}={(\frac{1}{2})}^{2}\) probability of winning on their first turn.

Alice has a \({(\frac{1}{2})}^{2}*\frac{1}{2}={(\frac{1}{2})}^{3}\) probability of winning on her their second turn.

Bob has a \({(\frac{1}{2})}^{3}*\frac{1}{2}={(\frac{1}{2})}^{4}\) probability of winning on their second turn.

Alice has a \({(\frac{1}{2})}^{4}*\frac{1}{2}={(\frac{1}{2})}^{5}\) probability of winning on her thrid turn.

We see Alice has a \({(\frac{1}{2})}^{1}+{(\frac{1}{2})}^{3}+{(\frac{1}{2})}^{5}+{(\frac{1}{2})}^{7}...\) chance of winning.

We can use the infinite geometric series fomula,\(\frac{\frac{1}{2}}{1-\frac{1}{4}}\), we get \(\frac{2}{3}\).

\(\)

Here is my method:

We simplify, 

\(f(x)={x}^{2}+14x\).

We might be tempted to substitute \(f(k-x)\), into the equation, but let's see if there's any other method.

We know \(f(k-x)\) represents a transformation.

First \(f(x+k)\) represents a translation left of k

Secondly, \(f(-x+k)\), represents a reflection across the y axis after the transformation.

First we know that the shape and size of the function after the transformation didn't change, because these transformations don't change the shape and size of an object.

But what we can do if calculate the shift of the quadratic through the roots. The original roots of the quadratic are 0 and -14.

Lets trace the path of the root 0, through this transformation.

We shift right left k, 0 becomes (-k, 0). Reflecting across the y axis, this becomes (k, 0). We know that (k, 0) is not the original root, but we see that for f(x) to map to f(k-x), then (k, 0) must represent the root (-14, 0)!!!

So, therefore k = -14.

Simplify to 

\(2{x}^{2}-14x+19\).

The general form of a quadratic is \(f(x) = a(x-x_{1})(x-x_{2}), f(x_{1})=0, f(x_{2})=0\), so if we want the same roots, we change a, by multiplying the quadratic by a number.

Because the constant term is 1, we divide by 19, also keeping the roots \(\frac{2}{19}{x}^{2}-\frac{14}{19}x+1\).

So A + B is \(\frac{2}{19}-\frac{14}{19}=-\frac{12}{19}\). 

\(4{x}^{2}+5x+k=0\).

Use discriminant so that \(\Delta = 0\).

\(25-4(k)(4)=0\)

\(16k=25\)

\(k=\frac{25}{16}\).