Math Is Fun Forum

In Euclidean geometry lines and points are drawn with magic ink*.  No matter the degree of magnification a point always looks the same, a dimensionless position. Worse than that, the magic ink is weightless so no weight device can determine a non zero weight for it.

Bob

* This ink is available from Bryan Thwaites' sale of useful equipment for schools teaching mechanics.  Other items include weightless pulleys, frictionless surfaces and non stretchable strings.

All correct.

Bob

I would do the same.  To get the boundary line it helps to have it in the form y = mx + c.  Then you just have to decide which side of the line satisfies the inequality.

The 'lazy' way to do that is to pick a point on one side and see if it fits the inequality.  As any point will do, choose one that makes the calculations easy.

Bob

I've got the book.

Bob

Ha ha! Nice answer.

Whilst most things in maths can be measured, it's an error to assume that {the set of things that cannot be measured} is empty.  Here's a few:

the focus of a parabola
the place where a curve meets its asymptote
the null set
the number of tesselating tiles needed to cover an infinite plane

In Euclidean geometry (which is where we came in) it is necessary that a point has no dimensions.  It exists because it is where two lines cross, but, if it had a size, that would mess up all the axioms.

Many years ago I did the "draw a triangle; measure the size of its angles; add them up; and consider what do you notice" with a class.  One student, who was clearly destined for great things, said "The angles add up to about 180, sometimes 179, sometimes 181."  Based on the class experiment this was a correct conclusion.  But Euclidean geometry requires that the angle sum of a triangle is exactly 180.  It doesn't work if you start worrying about the accuracy of the measurement.

Bob

I think my 'quadratic formula' method above is what you're seeking.

Here's why it works:

You're trying to factorise ax2 + bx + c.  Let's say it can be factored as (ax + p)(x + q)

Now consider the quadratic equation ax^2 + bx + c = 0.  If it can be factored you will have (ax + p)(x + q) = 0 so you will have found p and q for the trinomial.

Conversely if you can show the quadratic has no solution this means that p and q don't exist so the trinomial is prime.

Bob

Correct.

Bob

Correct.

Bob

Got it! well done.

Bob

That looks good.

Bob

Jumping into complex numbers is brilliant and a correct factorisation But the question says show prime ?  I think you weren't supposed to  go there but stick to real numbers. 

Have a look at P53 example 13

Proir to learning about complex numbers my teachers used to say 'cannot factorise' for questions like this.

Bob

x² + 4

Correct,

Bob