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Featured 3 months ago

Given:

#3a^2+4ab+b^2-2ac-c^2#

Note that all of the terms are of degree

So if this factors into simpler polynomials then its factors are homogeneous of degree

If we ignore the terms involving

#b^2-c^2 = (b-c)(b+c)#

If we ignore the terms involving

#3a^2+4ab+b^2 = (3a+b)(a+b)#

If we ignore the terms involving

#3a^2-2ac-c^2 = (3a+c)(a-c)#

These various linear binomial factors can be combined as follows:

#(b+c), (3a+b), (3a+c) rarr (3a+b+c)#

#(b-c), (a+b), (a-c) rarr (a+b-c)#

Hence we find the factorisation:

#3a^2+4ab+b^2-2ac-c^2 = (3a+b+c)(a+b-c)#

Featured 3 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 1 month ago

A way to get round the decimal place in the initial division expression.

Note that

Note that

So

I will do the division first then multiply by the

As we have 0 we may now stop the division

Now we multiply by the

~~~~~~~~~~~~~~~~~~~~~~~~~~~

If the division had not finished with a 0 at that point then we would have gone into decimal values.

Featured 1 month ago

see explanation.

The equation of a parabola in

#color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#

where (h ,k) are the coordinates of the vertex and a is a constant.

#y=2(x-3)^2+4" is in this form"#

#"with " a=2, h=3" and " k=4#

#rArrcolor(magenta)"vertex "=(3,4)#

#"since " a>0" then min. turning point "uuu#

#color(blue)"Intercepts"#

#â€¢ " let x = 0, in equation, for y-intercept"#

#â€¢ " let y = 0, in equation, for x-intercepts"#

#x=0toy=2(-3)^2+4=22larrcolor(red)" y-intercept"#

#y=0to2(x-3)^2+4=0#

#rArr(x-3)^2=-2# This has no real solutions hence f(x) has no x-intercepts.

graph{2(x-3)^2+4 [-31.56, 31.67, -15.8, 15.8]}

Featured 1 month ago

graph{2x-10 [-5, 10, -15, 5]}

(the graph has been scaled to show the intercepts and does not accurately depict the slope)

Put the equation into double-intercept form.

where

Original equation:

Divide all parts by 10:

And here is the finished product!

So the answer is...

Featured 4 days ago

We can begin by plotting the points of the line segment

The midpoint formula:

Since we have our two coordinates of

If we let,

Since we now know the coordinates of

By the same method above...

If we let,

So the coordinate of

Here is a graph to provide a visual representation of what I just explained.

If the image is to small, you can check the link below to this graph: https://www.desmos.com/calculator/vtaiir6ro8

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