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## Prove that if u is an odd integer, then the equation x^2+x-u=0 has no solution that is an integer?

Steve
Featured 5 months ago

Proposition
If $u$ is an odd integer, then the equation ${x}^{2} + x - u = 0$ has no solution that is an integer.

Proof
Suppose that there exists a integer solution $m$ of the equation:

${x}^{2} + x - u = 0$

where $u$ is an odd integer. We must examine the two possible cases:

$m$ is odd; or
$m$ is even.

First, let us consider the case where $m$ is odd, then there exists an integer $k$ such that:

$m = 2 k + 1$

Now, since $m$ is a root of our equation, it must be that:

${m}^{2} + m - u = 0$
 :. (2k + 1)^2 + (2k + 1) − u = 0
 :. (4k^2 + 4k + 1) + (2k + 1) − u = 0
 :. 4k^2 + 6k + 2 − u = 0
$\therefore u = 4 {k}^{2} + 6 k + 2$
$\therefore u = 2 \left(2 {k}^{2} + 3 k + 1\right)$

And we have a contradiction, as $2 \left(2 {k}^{2} + 3 k + 1\right)$ is even, but $u$ is odd.

Next, let us consider the case where $m$ is even, then there exists an integer $k$ such that:

$m = 2 k$

Similarly, since $m$ is a root of our equation, it must be that:

${m}^{2} + m - u = 0$
 :. (2k)^2 + (2k) − u = 0
 :. 4k^2 + 2k − u = 0
$\therefore u = 4 {k}^{2} + 2 k$
$\therefore u = 2 \left(2 {k}^{2} + k\right)$

And, again, we have a contradiction, as $2 \left(2 {k}^{2} + k\right)$ is even, but $u$ is odd.

So we have proved that there is no integer solution of the equation ${x}^{2} + x - u = 0$ where $u$ is an odd integer.

Hence the proposition is proved. QED

## How do you find the vertex and the intercepts for f(x) = -x^2 + 8x + 2?

Rachel
Featured 3 months ago

Vertex: $\left(4 , 18\right)$
Y-intercept: $\left(0 , 2\right)$
X-interecpt: $\left(4 + 3 \sqrt{2} , 0\right)$ and $\left(4 - 3 \sqrt{2} , 0\right)$

#### Explanation:

The vertex is pretty easy. We just need to complete the square .

First, the leading coefficient of the polynomial must be $1$, so we need to factor the $- 1$. That leaves us with $y = - 1 \left({x}^{2} - 8 x - 2\right)$. Now, the purpose of completing the square is to find a constant that makes ${x}^{2} - 8 x$ a perfect square. To that, we use this formula: $c = {\left(\frac{1}{2} \cdot b\right)}^{2}$ or ${\left(\frac{1}{2} \cdot 8\right)}^{2}$, which is $16$.

Now we know that we have to add $16$ to make it a pefect square, but because we cannot just add something on one side of the equation, we need to "get rid of it" too. We could add $16$ on both sides, or we can just add $16$ and then subtract it immediately, which is the same thing. Either way works :)
$y = - 1 \left({x}^{2} - 8 x \textcolor{g r e e n}{+ 16 - 16} - 2\right)$
$y = - \left(\left({x}^{2} - 8 x + 16\right) - 16 - - 2\right)$

${x}^{2} - 8 x + 16$ is a perfect square, so let's symplify it

$y = - \left({\left(x - 4\right)}^{2} - 16 - 2\right)$
$y = - \left({\left(x - 4\right)}^{2} - 18\right)$

Now we just distribute the negative:
$y = - {\left(x - 4\right)}^{2} + 18$

The equation is now in vertex form.

It's easy to find the vertex from this point:
$y = - {\left(x - \textcolor{red}{4}\right)}^{2} + \textcolor{p u r p \le}{18}$
$\left(\textcolor{red}{4} , \textcolor{p u r p \le}{18}\right)$.

Finding the $x$-interecpt means setting $y = 0$ and solving for $x$:

$0 = - {\left(x - 4\right)}^{2} + 18$
$- 18 = - {\left(x - 4\right)}^{2}$
$18 = {\left(x - 4\right)}^{2}$
$\pm \sqrt{18} = x - 4$
$4 \pm \sqrt{18} = x$
or $x = 4 \pm 3 \sqrt{2}$

Those are the exact values. If you want the estimated values, they're $x \approx 8.243$ and $x \approx - .0 .243$

To find the $y$-intercept we just set $x = 0$ and solve for $y$:
$y = - {\left(0 - 4\right)}^{2} + 18$
$y = - {\left(- 4\right)}^{2} + 18$
$y = - 16 + 18$
$y = 2$

## If f(x) = 3^x, what is f^-1(x)?

Featured 1 month ago

The inverse function is ${f}^{-} 1 \left(x\right) = \ln \frac{x}{\ln} 3$

#### Explanation:

Let $y = {3}^{x}$

To calculate the inverse, interchange $x$ and $y$,

$x = {3}^{y}$

Express $y$ in terms of $x$

$\ln x = \ln \left({3}^{y}\right)$

$\ln x = y \ln 3$

$y = \ln \frac{x}{\ln} 3$

Therefore,

${f}^{-} 1 \left(x\right) = \ln \frac{x}{\ln} 3$

Verification :

$f \left({f}^{-} 1 \left(x\right)\right) = f \left(\ln \frac{x}{\ln} 3\right) = {3}^{\ln \frac{x}{\ln} 3}$

Let $y = {3}^{\ln \frac{x}{\ln} 3}$

$\implies$

$\ln y = \ln \frac{x}{\ln} 3 \cdot \ln 3 = \ln x$

$y = x$

$f \left({f}^{-} 1 \left(x\right)\right) = x$

So, the functions ${3}^{x}$ and $\ln \frac{x}{\ln} 3$ are inverses

## How do you graph?? -2x+y=2

Mr. Raj
Featured 1 month ago

Use $y \text{-intercept}$ and $x \text{-intercept}$ or any two points on the line.

#### Explanation:

Looking at the equation, it is apparent that it would follow a linear model as there are no exponents or powers (eg ${x}^{2}$)

Then what we can do if find to points on the graph and draw a line through these points. The easiest points to find are usually the $y \text{-intercept}$ and $x \text{-intercept}$

The $y \text{-intercept}$ is when $x = 0$, so to find that point, make $x = 0$ in the equation, and solve for $y$

$x = 0$

$\implies - 2 \left(0\right) + y = 2$

$\implies y = 2$

Hence, when $x = 0$, $y = 2$. This represents the point $\left(0 , 2\right)$. So we should plot this point:

Then we calculate the other point, the $x \text{-intercept}$

The $x \text{-intercept}$ is when $y = 0$, so to find that point, make $y = 0$ in the equation, and solve for $x$

$y = 0$

$\implies - 2 x + \left(0\right) = 2$

$\implies - 2 x = 2$

$\implies x = \frac{2}{-} 2$

$\implies x = - 1$

Hence, when $y = 0$, $x = - 1$. This represents the point $\left(- 1 , 0\right)$. So we should plot this point with the other point:

Now join the two points with a straight line:

And you have the graph of your equation

You can use any two poinst on the line though. The only difference is that instead of finding when $x = 0$ or $y = 0$, you would find when $x = \text{another number like } 3$ or when $y = \text{another number like } 4$

It is easier to find the $y \text{-intercept}$ and $x \text{-intercept}$ but if you want to challenge yourself, find two points other than the $y \text{-intercept}$ and $x \text{-intercept}$

## How do you find the range of a function equations? I am so confused.

Vinay J.
Featured 4 weeks ago

Range of a $y = f \left(x\right)$ is a set of all the outputs corresponding to their input in the domain

#### Explanation:

there are many ways to find range
1. by constructing graph :
example
$y = \sin x$
Domain :: $\left(- \infty , \infty\right)$
Range :: $\left(- 1 , 1\right)$
because we get values between -1 and 1 (both included)
graph{sinx [-10, 10, -2, 2]}

1. solving $x = f \left(y\right)$
example :

$f \left(x\right) = \frac{2 x + 1}{x - 1}$
Domain : $R - \left\{1\right\}$

Now , forming equation such as $x = f \left(y\right)$

$x = \frac{y + 3}{y - 2}$
graph{(x +3)/(x -2) [-41.1, 41.08, -8.22, 8.22]}
Range : $R - \left\{2\right\}$

1. if you know that the function is continuous in domain then
then you can find max and min values of y

Ex . Range of $f \left(x\right) = {x}^{3} - 6 {x}^{2} + 11 x - 6$ where x lies b/w $\left(0 , 6\right)$
Ans. here, function is continuous in domain x : $\left(0 , 6\right)$
and ${y}_{\max} = f \left(6\right) = 60 \mathmr{and} {y}_{\min} = f \left(0\right) = - 6$
So, Range : $\left(- 6 , 60\right)$

Another Example which is important is when
$f \left(x\right) = {x}^{2} - 3 x + 2$ where x lies b/w $\left(- \infty , \infty\right)$
Here $f \left(x\right)$ is continuous in the domain and
${y}_{\max} = f \left(- \infty\right) = \infty$ but,
${y}_{\min} = f \left(\frac{3}{2}\right) = - \frac{1}{4}$
So,
Range is $\left[- \frac{1}{4} , \infty\right)$

## How do you find the slope that is perpendicular to the line 9x+3y=0?

Vinay J.
Featured 4 weeks ago

Condition :: ${m}_{1} {m}_{2} = - 1$
Slope :: $\frac{1}{3}$

#### Explanation:

Suppose that the slope of the 2 lines be ${m}_{1}$ and ${m}_{2}$ respectively
and angle between line and $x$ -axis be ${\theta}_{1}$ and ${\theta}_{2}$ respectively.

So, ${m}_{1} = \tan \left({\theta}_{1}\right) \mathmr{and} {m}_{2} = \tan \left({\theta}_{2}\right)$
we also know that ${\theta}_{1} - {\theta}_{2} = \frac{\pi}{2}$

then ,
$\tan \left({\theta}_{1} - {\theta}_{2}\right) = \infty$

but,

$\tan \left({\theta}_{1} - {\theta}_{2}\right) = \frac{\tan {\theta}_{1} - \tan {\theta}_{2}}{1 + \tan {\theta}_{1} \tan {\theta}_{2}}$

$\infty = \frac{{m}_{1} - {m}_{2}}{1 + {m}_{1} {m}_{2}}$

this can happen when

${m}_{1} {m}_{2} = - 1$
(this condition is very important )

substituting ${m}_{1} = - \frac{9}{3} = - 3$ (from the equation)
we get ${m}_{2} = \frac{1}{3}$

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