# Thread: [SOLVED] Help Needed...

1. ## [SOLVED] Help Needed...

-Show that x-a is a factor of x^n-a^n for any positive integer n.

-Show that x+a is a factor of x^n+a^n for any positive odd integer n.

Prove that if a≠b, then (a, b) and (b, a) lie on a line perpendicular to the line y=x and are equidistant from y=x.

Prove that if f is a one-to-one odd function, f-^-1 is an odd function.

Consider the following system of equations in x+y:

ax+by=c
dx+ey=f

Where the constants a, b, c, d, e, and f form an arithmetic sequence. How might you characterize the solutions to the system of equations?

2. Originally Posted by Szpieg
-Prove that if a≠b, then (a, b) and (b, a) lie on a line perpendicular to the line y=x and are equidistant from y=x.
The slope of the line y = x is 1. The slope of the line with the points (a, b) and (b, a) is
$\frac{y_2 - y_1}{x_2 - x_1} = \frac{a - b}{b - a} = -1$ ,
so this line is perpendicular to the line y = x. (Slopes of perpendicular lines are negative reciprocals of each other.)

Pick a point (c, c) that is on the line y = x. Use the distance formula to find the distance from (a, b) to (c, c), and from (b, a) to (c, c).

01

3. thanks for the help, if anyone can help we with the others, it would be much appreciated...

4. Originally Posted by Szpieg
-Consider the following system of equations in x+y:

ax+by=c
dx+ey=f

Where the constants a, b, c, d, e, and f form an arithmetic sequence. How might you characterize the solutions to the system of equations?
If the constants a, b, c, d, e, f form an arithmetic sequence, then
b = a + n
c = a + 2n
d = a + 3n
e = a + 4n
f = a + 5n
for some n. Substitute:

\begin{aligned}
ax\ &+ &(a + n)y\ &=\ a + 2n \\
(a + 3n)x\ &+ &(a + 4n)y\ &=\ a + 5n
\end{aligned}

Solve by elimination. Multiply the 1st equation by -(a + 3n)/a:

\begin{aligned}
-(a + 3n)x\ &- &\frac{(a + n)(a + 3n)}{a}y\ &=\ -\frac{(a + 2n)(a + 3n)}{a} \\
(a + 3n)x\ &+ &(a + 4n)y\ &=\ a + 5n
\end{aligned}

Add the 2 equations to cancel out the x's. The y coefficients, added together, give us

$- \frac{(a + n)(a + 3n)}{a} + (a + 4n) = - \frac{a^2 + 4an + 3n^2}{a} + \frac{a^2 + 4an}{a} = -\frac{3n^2}{a}$ ,

and the constant terms, added together, give us

$- \frac{(a + 2n)(a + 3n)}{a} + (a + 5n) = - \frac{a^2 + 5an + 6n^2}{a} + \frac{a^2 + 5an}{a} = -\frac{6n^2}{a}$ .

The two equations added together, therefore, is

$-\frac{3n^2}{a}y = -\frac{6n^2}{a}$ ,

meaning that y = 2.

Substitute y = 2 into either of the original equations (I chose the first):

\begin{aligned}
ax + (a + n)y &= a + 2n \\
ax + (a + n)(2) &= a + 2n \\
ax + 2a + 2n &= a + 2n \\
ax &= -a
\end{aligned}

So x = -1. We therefore have one solution to the system of equations, (-1, 2).

01

5. OMG, thank you so much, this one i had no idea how to do=] ull be getting thanks from me =] and respect...by the way how old are you and what school do you go to?

PS, i got the same answer, but not in a true proof...i used multiple examples, but my teacher does not want that...this is perfect

6. Originally Posted by Szpieg
-Show that x-a is a factor of x^n-a^n for any positive integer n.

-Show that x+a is a factor of x^n+a^n for any positive odd integer n.

-Prove that if a≠b, then (a, b) and (b, a) lie on a line perpendicular to the line y=x and are equidistant from y=x.

-Prove that if f is a one-to-one odd function, f-^-1 is an odd function.

-Consider the following system of equations in x+y:

ax+by=c
dx+ey=f

Where the constants a, b, c, d, e, and f form an arithmetic sequence. How might you characterize the solutions to the system of equations?
For your first question, call your function $f(x)$.

So $f(x) = x^n - a^n$.

By the Factor and Remainder Theorems, $(x - a)$ is a factor IF $f(a) = 0$.

So $f(a) = a^n - a^n = 0$.

So $(x - a)$ is a factor of $x^n - a^n$.

An alternative proof is done using factorisation.

For any $a, b$

$a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \dots + a^2b^{n - 3} + ab^{n - 2} + b^{n - 1})$.

Applying this knowledge to $f(x)$ we find

$f(x) = (x - a)(x^{n - 1} + x^{n - 2}a + x^{n -3}a^2 + \dots + x^2a^{n - 3} + xa^{n - 2} + a^{n - 1})$

So it can clearly be seen that $(x - a)$ is a factor.

For your second problem, to show $(x + a)$ is a factor of $x^n + a^n$ for any $n$ that is an odd positive integer, do the same thing.

Let $f(x) = x^n + a^n$.

By the Factor and Remainder Theorems,
$(x + a)$ is a factor if $f(-a) = 0$.

So $f(-a) = (-a)^n + a^n$.

If $n$ is even, you have $f(-a) = a^n + a^n = 2a^n$, so $x - a$ is not a factor if $n$ is even,

BUT If $n$ is ODD, you have $f(-a) = -a^n + a^n = 0$.

So $(x + a)$ is a factor of $f(x)$ as long as n is a positive odd integer.

Alternatively, use a similar alternate proof to the one above.

If $n$ is an odd integer, then

$a^n + b^n = (a + b)(a^{n - 1} - a^{n - 2}b + a^{n - 3}b^2 - \dots + a^2b^{n - 3} - ab^{n - 2} + b^{n - 1})$.

So $f(x) = x^n + a^n = (x + a)(x^{n - 1} - x^{n - 2}a + x^{n - 3}a^2 - \dots + x^2a^{n - 3} - xa^{n - 2} + a^{n-1})$.

Here you can clearly see that $x + a$ is a factor of $f(x)$, but only if $n$ is a positive odd integer.[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG]