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Featured 5 months ago

**Proposition**

If

**Proof**

Suppose that there exists a integer solution

#x^2 + x - u = 0#

where

#m# is odd; or

#m# is even.

First, let us consider the case where

# m = 2k + 1 #

Now, since

# m^2 + m - u = 0 #

# :. (2k + 1)^2 + (2k + 1) âˆ’ u = 0 #

# :. (4k^2 + 4k + 1) + (2k + 1) âˆ’ u = 0 #

# :. 4k^2 + 6k + 2 âˆ’ u = 0 #

# :. u = 4k^2 + 6k + 2 #

# :. u = 2(2k^2 + 3k + 1) #

And we have a contradiction, as

Next, let us consider the case where

# m = 2k #

Similarly, since

# m^2 + m - u = 0 #

# :. (2k)^2 + (2k) âˆ’ u = 0 #

# :. 4k^2 + 2k âˆ’ u = 0 #

# :. u = 4k^2 + 2k #

# :. u = 2(2k^2 + k) #

And, again, we have a contradiction, as

So we have proved that there is no integer solution of the equation

Hence the proposition is proved. QED

Featured 3 months ago

Vertex:

Y-intercept:

X-interecpt:

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

First, the leading coefficient of the polynomial must be

Now we know that we have to add

Now we just distribute the negative:

The equation is now in vertex form.

It's easy to find the vertex from this point:

Finding the

or

Those are the exact values. If you want the estimated values, they're

To find the

Featured 1 month ago

The inverse function is

Let

To calculate the inverse, interchange

Express

Therefore,

Verification :

Let

So, the functions

Featured 1 month ago

Use

Looking at the equation, it is apparent that it would follow a linear model as there are no exponents or powers (eg

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

The

Hence, when

Then we calculate the other point, the

The

Hence, when

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

It is easier to find the

Featured 4 weeks ago

Range of a

there are many ways to find range

1. by constructing graph :

example

Domain ::

Range ::

because we get values between -1 and 1 (both included)

graph{sinx [-10, 10, -2, 2]}

- solving
#x = f(y) #

example :

Domain :

Now , forming equation such as

graph{(x +3)/(x -2) [-41.1, 41.08, -8.22, 8.22]}

Range :

- if you know that the function is continuous in domain then

then you can find max and min values of y

Ex . Range of

Ans. here, function is continuous in domain x :

and

So, Range :

Another Example which is important is when

Here

So,

Range is

Featured 4 weeks ago

Condition ::

Slope ::

Suppose that the slope of the 2 lines be

and angle between line and

So,

we also know that

then ,

but,

this can happen when

(this condition is very important )

substituting

we get

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