Wikipedia:Reference desk/Archives/Mathematics/2017 August 24

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August 24

Resources to learn A-level (year 11 & 12) maths?

I'll be doing A-level maths independently and am looking for additional resources for when the textbook isn't enough to explain the problem. 2.102.187.156 (talk) 12:56, 24 August 2017 (UTC)

There are several websites such as this, this, this, this, this this this this this and this, many of them free. I haven't checked to see which is best. Most seem to follow the Edexcel syllabus. You can always ask here if you have specific problems you are having difficulty with. Dbfirs 13:36, 24 August 2017 (UTC)

Khan Academy is a good resource; it's more geared towards the US Curriculum in terms of pacing and order of subjects, but if you know specific subject matter, you can follow the videos there as an additional resources. here is their library of all topics covered; if you have specific topics on the A-levels you want additional help with, that could be an additional resource. --Jayron32 14:11, 24 August 2017 (UTC)

Yes, that's an excellent resource if you pick the appropriate topics from your syllabus. There are some minor differences in terminology (such as plotting on the "complex plane" instead of an "Argand diagram"), and you will need to be careful to choose the right level, but the explanations are better than I've seen on any British website. Dbfirs 16:07, 24 August 2017 (UTC)

Cheers those are really helpful. 2.102.187.156 (talk) 23:48, 26 August 2017 (UTC)

Simultaneous quadratic equations

I have three quadratic equations with three unknowns (x,y,z). What is the easiest way to solve for x,y,z?

(x-a1)^2 + (y-b1)^2 =(z-c1)^2
(x-a2)^2 + (y-b2)^2 =(z-c2)^2
(x-a3)^2 + (y-b3)^2 =(z-c3)^2

-- SGBailey (talk) 15:22, 24 August 2017 (UTC)

Perform the binomial expansions, collect like terms, then solve as a system of equations. --Jayron32 15:35, 24 August 2017 (UTC)

Haven't looked to solving it, but I can say that the locus of a point where the distance to two other points differs by s constant is a hyperbola. And two hyperbolas can meet at up to four points. Dmcq (talk) 15:39, 24 August 2017 (UTC)

This actually looks like an intersection of three cones with different coordinate origins. To solve it just subtract the second equation from the first then the third from the first and finally the third from the second. As a result you will get a system of three linear equations:

$(a2-a1)(x-a1-a2)+(b2-b1)(y-b1-b2)=(c2-c1)(z-c1-c2)$

$(a3-a1)(x-a1-a3)+(b3-b1)(y-b1-b3)=(c3-c1)(z-c1-c3)$

$(a3-a2)(x-a2-a3)+(b3-b2)(y-b2-b3)=(c3-c2)(z-c2-c3)$

Only two equations are independent

$(a2-a1)x+(b2-b1)y-(c2-c1)z=a2^{2}-a1^{2}+b2^{2}-b1^{2}-c2^{2}+c1^{2}$

$(a3-a1)x+(b3-b1)y-(c3-c1)z=a3^{2}-a1^{2}+b3^{2}-b1^{2}-c3^{2}+c1^{2}$

You need to express two variable through the third then substitute these expressions into one of the original equations and solve the resulting quadratic equation. Ruslik_Zero 17:56, 24 August 2017 (UTC)

Nice approach, Ruslik. But I would add a caveat. It's true that the left hand side matrix of your linear system has a zero determinant, meaning that if the three equations are consistent then one of them is redundant. But the third equation could be inconsistent with the other two due to its right hand side constant, meaning that the linear system and thus the original quadratic system has no solution. The can be checked by considering the rank of the augmented matrix—if it is greater than the rank of the unaugmented matrix, the linear system is inconsistent. Loraof (talk) 19:37, 24 August 2017 (UTC)

The third equation can be obtained by subtracting the first equation from the second. So, it adds no new constants on the solution. Therefore, the system is always consistent. Ruslik_Zero 20:18, 24 August 2017 (UTC)

Thanks all - I'll give it a go. -- SGBailey (talk)

Not true, Ruslik. For a trivial example, let c1=1, c2=2, c3=3, and all other parameters =0. Then x^2+y^2=(z-1)^2, x^2+y^2=(z-2)^2, and x^2+y^2=(z-3)^2, which is inconsistent. Your corresponding linear equations are 0=z–3, 0=2(z–4), and 0=z–5. Loraof (talk) 21:42, 24 August 2017 (UTC)

• Further, we can say that almost all parametrizations are inconsistent. Using the left side matrix and augmented matrix I referred to earlier, the left side matrix always has zero determinant, so its rank is less than or equal to 2, and will generally be 2. But the augmented matrix does not in general have rank 2; if you delete one of the first three columns from the augmented matrix (a redundant column), the rank of the resulting 3×3 matrix is the same as the rank of the augmented matrix; for this to be 2 requires that the determinant of the latter 3×3 matrix be 0, which imposes a restriction on the relation of the parameters to each other. For randomly chosen parameters this relation will almost surely not hold.

So using Ruslik's approach gives candidate solutions for the original quadratic system, but in general these will not be valid solutions and have to be checked against all three quadratic equations. Loraof (talk) 22:06, 24 August 2017 (UTC)

Loralof, I don't know what you're talking about: in the relevant 3 by 3 linear system, the third equation is *always* consistent with the other two, just as Ruslik0 says. This is clear both from the equations and from the process by which they were constructed. --JBL (talk) 12:05, 25 August 2017 (UTC)

Then what's a solution to the case I mentioned: c1=1, c2=2, c3=3, all other parameters=0, which gives the quadratic system $x^{2}+y^{2}=(z-1)^{2}=(z-2)^{2}=(z-3)^{2},$ and what's the solution to its linear counterpart $0=z-3,\,0=2(z-4),\,0=z-5$? Loraof (talk) 15:27, 25 August 2017 (UTC)

The ranks of the left side and augmented matrices are at most 2, not 3. This follows from the fact that you can eliminate one of the equations. Then the left side matrix is 2×3 and the augmented matrix is 2×4. So, their ranks can not exceed 2. The pathological example that you found (and which, I admit, I missed) corresponds to case when the left side matrix has rank 1. So, in this case there is indeed no solution. Ruslik_Zero 17:55, 25 August 2017 (UTC)

Loralof, the answer to your question is that the third equation in that case is implied by the other two, which are inconsistent with each other. --JBL (talk) 01:52, 26 August 2017 (UTC)

The 3 equations are

0 = $(x-a_{i})^{2}+(y-b_{i})^{2}-(z-c_{i})^{2}$ for $i=1,2,3$.

Expand the squares

0 = $x^{2}-2a_{i}x+a_{i}^{2}+y^{2}-2b_{i}y+b_{i}^{2}-z^{2}+2c_{i}z-c_{i}^{2}$

Subtract the first one from the second one and from the third one

0 = $A_{i}x+B_{i}y-C_{i}z+D_{i}$ for $i=2,3$

where

$A_{i}=2(a_{1}-a_{i})$

$B_{i}=2(b_{1}-b_{i})$

$C_{i}=2(c_{1}-c_{i})$

$D_{i}=-a_{1}^{2}+a_{i}^{2}-b_{1}^{2}+b_{i}^{2}+c_{1}^{2}-c_{i}^{2}$

Now the squares have been eliminated from the second and the third, but not from the first equation. Multiply the first one by $A_{2}$ and the second one by $x$ and subtract in order to eliminate $x^{2}$

0 = $Ex+F$

where

$E=-2A_{2}a_{1}-B_{2}y+C_{2}z-D_{2}$

and

$F=A_{2}y^{2}-2A_{2}b_{1}y-A_{2}z^{2}+2A_{2}c_{1}z+A_{2}b_{1}^{2}-A_{2}c_{1}^{2}$

This equation, and equations 2 and 3, are of degree 1 in $x$ . So $x$ can be eliminated from equations 2 and 3, and we are left with 2 equations in the 2 variables $y$ and $z$

0 = $E(B_{i}y-C_{i}z+D_{i})-A_{i}F$ for $i=2,3$

Continuing this procedure eliminate first $y^{2}$, and then $y$, to obtain one algebraic equation in the variable $z$. Bo Jacoby (talk) 16:39, 25 August 2017 (UTC).

Diophantine question...

Does x/(y+z)+y/(x+z)+z/(x+y) = N have a solution where x,y,z and N are all integers?— Preceding unsigned comment added by Naraht (talk • contribs)

Yes. Many solutions, such as x=1, y=1, z=3. That would be 1/4+1/4+3/2=2. 209.149.113.5 (talk) 16:40, 24 August 2017 (UTC)

And of course, by symmetry of the original equation, any permutation of a solution is a solution also. And also kind of trivially, any multiple of a solution will be a solution as well (like (2,2,6), (3,3,9), etc.). --Deacon Vorbis (talk) 16:44, 24 August 2017 (UTC)

OK, so we have k*(1,1,3),2 and k*(1,1,0),2. Any more come to mind?

Sure, e.g. (-2,1,1) and (-5,1,1), both for N=-3. There are probably others. --Deacon Vorbis (talk) 17:18, 24 August 2017 (UTC)

I think those may be the only solutions with all non-negative integers. If you consider negative integers as well, then there are many additional solutions. For example for every set of positive integers (a,b,k) there is a solution k*(a,-b,b-a), -3. Also, k*(-5,6,7), 6; k*(-2,1,5), -5; k*(-7,-5,9), -5; k*(-7,-5,8), 8; k*(-5,4,41), -41; etc. Dragons flight (talk) 17:59, 24 August 2017 (UTC)

Another general class for every positive integer a is k*(-2a^2-2a-1, -a, a+1), -2a^2-2a-1. Dragons flight (talk) 18:09, 24 August 2017 (UTC)