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Correct.
Bob
I would expect 4 terms for that.
3^2 = 9 4^2 = 16 2x3x4 = 24,
Try a direct factorisation with two identical factors.
Bob
Yes.
Bob
Good so far. Sum of cubes? Try it.
Bob
The Merchant of Venice - Portia's speech, script by Shakespeare
Bob
The quality of mercy is not strained.
Bob
hi Irene,
Sometimes a quadratic equation will factorise
eg.
But often they don't. There's an algebraic 'trick' that allows you to get the solutions when this happens. It is called 'completing the square'.
eg.
This time there is no easy factorisation.
Step 0ne. Shift the number to the other side:
Step Two. Add (half the x coefficient) squared to both sides:
This is guaranteed to make the left hand side a perfect square .... it can be factorised as something^2
Step Three. Factorise the LHS
Step Four. Square root both sides.
Note: Two possible square roots unless that term is zero.
Step Five. Shift that non-x term to the RHS
Many years ago when I was taking my exams we were told we could use that method or use 'the formula' For a long time I did because the formula looked too complicated for me.
So you're not alone.
The formula is derived using completing the square, but with letters rather than specific numbers. Here we go ![]()
Step One. Divide all by 'a' so the first term is x^2
Note: 0 divided by a is still 0
Step Two. Move the non-x term to the other side.
Step Three. Add (half the x coefficient) squared to both sides:
Step Four. Simplify the RHS a bit.
Step Five. Square root both sides
Step Six. Move the non x term to the other side
Step Seven. Put all over the common denominator and 'loose' the unnecessary + sign
Bob
They are the same but what did the question say? Sometimes you're asked to factorise fully. In an exam, read the question carefully to make sure of this point. If it doesn't say then either should get full marks.
Some years ago I did some A level marking for Ed-Excel. The first question asked candidates to do a certain bit of algebra (part A) and then to use the result to do something else (part B). Other ways of doing this second part could have been used except for that crucial wording. The chief examiner told us to give no marks for part B to any candidate who didn't use part A. Along with other markers I was very unhappy about this but the chief's word is law so I did as told.
You'll be pleased to know that the standard of marking is very high. The mark scheme shows all the alternative ways that may earn marks and every bit gets working credit. If we couldn't read a candidate's writing or had some other problem with awarding marks we had to send it on to our superviser, who also checked a sample of our marking.
I have to say, I only did it that once. It was very tedious and not how I wanted to spend my life. Useful insight though.
Bob
(x - 1)(x + 1)(x^2 + 1)
Correct!
(x - 1)^2(x^2 + 1)
This is not the same. It is (x-1)(x-1))x^2+1)
Bob
Nothing else can be done here.
3 linear factors! I'd say that is fully done.
Bob
Correct.
Bob
That's it done nicely.
Bob
Yes, what you have done is correct.
Note: The question did not say to solve for p.
That's because the original was p^2 + 14p but it wasn't equal to anything so there wasn't actually an equation to solve.
Bob
Correct!
Bob
Let 27 be 3^3. This means a = 3 in the formula.
Let 8x^3 be (2x)^3.
Good
But then
(2x - 3)[(2x)^2 - 3(2x) + 3^2]
Bob
Ok. First notice that putting x=a causes x^5 - a^5 to become 0, so (x-a) is a factor => no remainder.
x^4 + a x^3 + a^2 x^2 + a^3 x + a^4
_____________________________________________________
x - a | x^5 + 0 x^4 + 0 x^3 + 0 x^2 + 0 x - a^5
x^5 - a x^4
______________
a x^4 + 0 x^3
a x^4 -a^2 x^3
___________________
+ a^2 x^3 + 0 x^2
+ a^2 x^3 - a^3 x^2
_________________________
+ a^3 x^2 + 0 x
+ a^3 x^2 - a^4 x
_____________________
+ a^4 x - a^5
+ a^4 x - a^5
_______________
remainder = 0 0
so x^5 - a^5 = ( x^4 + a x^3 + a^2 x^2 + a^3 x + a^4)(x-a)
Bob
Easier .... like using the quadratic formula rather than completing the square. But where did the formula come from? Mathematicians like to know the answer to that question ... understanding rather than rote learning.
Bob
Not me. I did wonder about them though. I count 6 missing. This forum has members from all over the world. That means a spread of politics and religion. All posters need to think about this and take care with potentially controversial subjects. But we do allow a wide range of topics so I was inclined to wait and see. We had a poster a while back who started with an innocent statement but it provoked unpleasant responses so in the end we had to terminate it.
I recommend you stick to maths.
Bob
I get what you're saying but I don't think so. From the wording b and c have values we can compute whereas x is truely a variable in the sense that, once you have computed b and c, the equation has to hold true whatever value x takes.
Like the circle question this is something I've always accepted and used without questioning it. I like that you do question it; that's the mark of a true mathematician. So now I'll have to think up a proof for paragraph 1.
LATER EDIT:
Substitute some values:
x = 1 => b = 3^c .........equation a
x = 2 => 64b= (3 times 4)^c = 3^c times 4^c .........equation b
Substitute a into b => 64b = b times 4^c => 4^c = 64 => c = 3
=> b = 27
Bob
Where did that question spring from.
If r = 0.546189584... then a sphere's surface area = 4, so you cannot rule that out.
A cuboid 1 by 1 by 0.5 has surface area 4.
A cube with side 0.816497.... has surface area 4.
There are loads of possible shapes so there's not a lot you can say for certain. A sphere has minimal surface area so you can probably say at least 1 dimension will be over the r given above. You can probably get an upper bound for the shapes dimensions and I think that's about it.
A solid shape could certainly be a compound shape.
Bob
The idea comes from Euclid. He wrote out a set of geometry axioms for an idealised version of geometry. He then showed how to derive many theorems from these and wrote a book "The Elements" that has been used by mathematicians ever since.
Not only does it give us many famous results (eg the angles of a triangle add up to 180) but his axiomatic approach has given mathematicians the way to develop any mathematical theory.
He defined a point as having a position but no size. Lines have length but no width. It's an idealised view. When you mark a point on paper and look at it under magnification you can see your point does have size.
Bob
When I did my age 18 A levels, stats wasn't part of the course.
But it was introduced later and I had to teach it. Sometimes I have to look up the methods.
Bob
This question doesn't need calculus. It does make use of the rule that if m1 times m2 = -1 for gradients of two lines, they are perpendicular.
Bob
p44 example 10 has (x-1). But the quesion here shows this can be generalised.
Bob
If this is question 37 then you just need the area of the circle ie pi
Bob