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**pi_cubed****Member**- From: A rhombicosidodecahedron
- Registered: 2020-06-22
- Posts: 114

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I agree with Alon Amit. This saddens me a little

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 38,747

In my opinion, the solution of a quadratic equation

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**pi_cubed****Member**- From: A rhombicosidodecahedron
- Registered: 2020-06-22
- Posts: 114

For some reason this has double posted; could an admin delete the second one?

pi³

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 38,747

Done.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 698

pi_cubed wrote:

I think this method is somewhat similar to factorization method for

Solving quadratic equations.

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,533

If you 'complete the square' or use the formula you have to do some calculations involving c and b squared, do a square root, add in -b and divide by a term with a in it.

This 'new' method involves eliminating a from the expression first, then a calculation involving b squared and c, then square rooting and then adding in -b/2.

So in what sense is this quicker, or different from what we've all been doing all these years? If you learn an algorithm and learn to do it quickly, then that's fine for you. But is this really worth all the hype?

later edit:

quadratic formula when a = 1 :

So this 'new' method is just the old one with a = 1.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,533

Quadratic Equations

There has been a lot of internet ‘fuss’ generated by Po-Shen Loh’s ‘new’ method for solving quadratic equations. One thing to say about that. It’s definitely not new. I spent a few minutes looking at the comments from his Utube and found the following:

“In Indian schools this is taught to students in grade 8 as a standard method”

“I was taught this method by my high-school math teacher in former Soviet Union in early nineties”

“My geometry teacher taught me this 3 years ago”

“I already known that I learned that in middle school”

“This is the thing i had already been taught in 9th class”

So, not new then.

I’m not sure if he is also claiming the sum and product of the roots bit but that’s certainly not new. I was taught that in 1967 as part of my A level and it was in the text that had been around for years.

I thought I would write up the standard quadratic theory and compare. Apologises to those for whom this is ‘old hat’ but there will be some readers who may find this useful.

A quadratic is an equation of the form ax^2 + bx + c = 0 for values of a, b and c. Apart from in mathematics they are used in physics, economics, astronomy, engineering, agriculture, military and law enforcement to mention a few.

If we draw the graph of y = ax^2 + bx + c we get something like this:

Using calculus dy/dx = 2ax + b and so the stationary point is when x = -b/(2a)

The gradient graph is linear and crosses the x axis at that point. If we differentiate again we get d2y/dx^2 = 2a. When a is positive this is positive so the gradient line goes from negative to positive values as we pass from left to right through the stationary point. So it is a minimum. If a is negative we get a maximum.

With k>0, if we consider a point to the right of the minimum where x = -b/(2a) + k then

And to the left, with x = -b/(2a) - k

These y values are the same. As this is true for all values of k this makes the vertical line x = -b/(2a) a line of symmetry for the graph. All quadratics are symmetrical in this way.

So what value(s) of x will fit the equation? The usual thing we do to solve an equation is to get x on the left hand side and numbers on the right and then calculate the right. But x occurs twice in a quadratic once as a square and that messes up simple manipulation of the equation. x^2 = expression with x in and x = expression with x^2 in are no help so something else is needed.

One way is to complete the square. I’ll illustrate with 3x^2 + 2x -20 = 0

Step 1. Divide through by 3 x^2 + 2x/3 -20/3 = 0

Step 2. Move the constant term over to the right and add (half the coefficient of x) squared to both sides.

x^2 + 2x/3 + 1/9 = 20/3 + 1/9

Step 3.

This makes the LHS a ‘perfect square’ so it will factorise with one repeated bracket.

(x + 1/3)^2 = 20/3 + 1/9

Step 4. Note that x only occurs once now so we can proceed to make it the subject of the equation.

Take the square root of both sides, remembering that a number has two square roots (+ and -)

x + 1/3 = +/- root(20/3 + 1/9)

Step 5. Subtract 1/3 from each side to finish the task.

If you apply this same process to the form with a, b and c in it you get:

That’s called the quadratic formula. You can still use ‘completing the square’ if you prefer. It’s essentially the same steps and the same amount of calculation … just means you do several small steps instead of plugging the whole thing into a formula, which you have to learn.

In Po-Shen Loh’s original article (https://arxiv.org/pdf/1910.06709.pdf) he uses an example where a=1. This ‘cherry picking’ of a favourable example is hardly a rigorous way to introduce his method.

If we put a = 1 in the formula we get:

Splitting this into 2 terms and moving the ‘over2’ inside the square root we get

which is his formula. He says his method makes it unnecessary to memorize any formula at all, but here it is in a screen shot I’ve taken from his article:

Is it a quicker and simpler way to solve a quadratic? Well given that you can only use his method by dividing through by ‘a’, and you still have to do a calculation with b^2 and c, and square root and do the +/- bit, it seems pretty much the same to me. It’ll work, so use it if you want. Computationally it comes to exactly the same thing. Well of course it does. It would be rather worrying if you got a different answer wouldn’t it?

Best wishes, keep safe!

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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