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Answer:

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

Explanation:

Given:

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

Note that all of the terms are of degree #2#.

So if this factors into simpler polynomials then its factors are homogeneous of degree #1#.

If we ignore the terms involving #a#, then we are looking for a factorisation of #b^2-c^2#, which can be written:

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

If we ignore the terms involving #c#, then we find a factorisation:

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

If we ignore the terms involving #b#, then we find a factorisation:

#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)#

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 = 2k + 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 #
# :. u = 4k^2 + 6k + 2 #
# :. u = 2(2k^2 + 3k + 1) #

And we have a contradiction, as #2(2k^2 + 3k + 1)# 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 = 2k #

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

And, again, we have a contradiction, as #2(2k^2 + k)# 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

Answer:

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

#159.9#

Explanation:

Note that #9.594# is the same as #9594xx1/1000#

Note that #0.06# is the same as #6xx1/100#

So #9.594-:0.006# is the same as #[9594-:6]xx[1/1000-:1/100]#

#(9594-:6)xx1/10#

I will do the division first then multiply by the #1/10# at the very end

#" "9594#
#color(magenta)(1000)xx6->ul(6000) larr" subtract"#
#" "3594#
#color(magenta)(color(white)(1)500)xx6->ul(3000) larr" subtract"#
#" "594#
#color(magenta)(color(white)(10)90)xx6->ul(color(white)(0)540)larr" subtract"#
#" "color(white)(0)54#
#color(magenta)(color(white)(100)9)xx6->" "ul(54) larr" subtract"#
#" "0#

As we have 0 we may now stop the division

#9594-:6 = color(magenta)(1599)#

Now we multiply by the #1/10# giving:

#159.9#
~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

Answer:

see explanation.

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"#

#• " let x = 0, in equation, for y-intercept"#

#• " 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]}

Answer:

#x=5 and y=-10#
graph{2x-10 [-5, 10, -15, 5]}
(the graph has been scaled to show the intercepts and does not accurately depict the slope)

Explanation:

Put the equation into double-intercept form.
#(x/a)+(y/b)=1#
where #a# is the x-intercept and #b# is the y-intercept

Original equation:
#2x-y=10#
Divide all parts by 10:
#((2x)/10)+((-1y)/10)=(10/10)#

And here is the finished product!
#(x/5)+(y/-10)=1#

So the answer is...
#x=5 and y=-10#

Answer:

#(11,7)#

Explanation:

We can begin by plotting the points of the line segment #PS#. Since we know that #R# is the midpoint of the #PS# line segment, we can use the midpoint formula to find the coordinate of #R#

The midpoint formula: #((x_1+x_2)/2, (y_1+y_2)/2)#

Since we have our two coordinates of #P# and #S# we can substitute those values into the formula to find the coordinates of #R#

If we let, #(8,10)->(color(blue)(x_1),color(red)(y_1))# and #(12,6)->(color(blue)x_2,color(red)(y_2))# then.. .

#R= (color(blue)((8+12)/2), color(red)((10+6)/2))=(color(blue)(20/2),color(red)(16/2))=(color(blue)10,color(red)8)#

Since we now know the coordinates of #R# and already given #S# we can find the coordinate of #Q# since it is the midpoint of #RS#

By the same method above...

If we let, #(10,8)->(color(blue)(x_1),color(red)(y_1))# and #(12,6)->(color(blue)x_2,color(red)(y_2))# then.. .

#Q=(color(blue)((10+12)/2), color(red)((8+6)/2))=(color(blue)(22/2),color(red)(14/2))=(color(blue)11,color(red)7)#

So the coordinate of #Q# lies at #(11,7)# which is also the midpoint of line segment #RS#

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

enter image source here

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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