Anyone good with math around?

[TheoG]

There are always many different ways to write the steps of the same problem but luckily the first/basic principles always remains the same.

[Orbit]

Eish, where to begin.

 

1. These are equations NOT expressions - that means that to solve for x you must use inverse operations. By extension, this means that your order of operations will be reversed.

Instead of doing Brackets - Exponents - Division/multiplication - Addition/subtraction (BEDMAS) you must do Subtraction/addition - Multiplication/Division - Exponents - Brackets (SAMDEB).

 

2. The first example is a Quadratic (x^2) equation and the second a quartic (x^4) equation but they both behave the same. Since x has an even power, it will have 2 solutions (one +ve and one -ve) hence you cannot solve by "getting x on its own" and square rooting as you will "lose a solution".

 

You MUST use the method of difference of 2 squares.

ie.

x^2 - b^2 = 0      NB to get everything = 0 with a quadratic eqn. 

(x + b)(x - b) = 0

x + b = 0 Or x - b = 0

x = -b Or x = b

[Orbit]

1.

(5x^4)/3 - 20 = 0

5x^4 - 60 = 0

Divide everything by 5 and factorise

 

x^4 - 12 = 0

(x^2 - sqt12)(x^2 + sqtt12) = 0

x^2 - sqt12 = 0. OR x^2 + sqt12 = 0

Now you must factorise  the x^2 - sqt12 = 0 again

 

This is difficult to write on computer but you will essentially get x = +//12 and x = - //12

 

2. 

(3x)^2 +36^1/2 = 100

9x^2 + 6 = 100

9x^2 - 94 = 0

Now factorise

 

(3x - sqt94)(3x + sqt94) = 0

3x - sqt94 = 0 OR 3x + sqtt 94 = 0

x = (sqt93)/3. Or x = - (sqt94)/3

[mecheng89]

Eish, where to begin.

 

1. These are equations NOT expressions - that means that to solve for x you must use inverse operations. By extension, this means that your order of operations will be reversed.

Instead of doing Brackets - Exponents - Division/multiplication - Addition/subtraction (BEDMAS) you must do Subtraction/addition - Multiplication/Division - Exponents - Brackets (SAMDEB).

 

2. The first example is a Quadratic (x^2) equation and the second a quartic (x^4) equation but they both behave the same. Since x has an even power, it will have 2 solutions (one +ve and one -ve) hence you cannot solve by "getting x on its own" and square rooting as you will "lose a solution".

 

You MUST use the method of difference of 2 squares.

ie.

x^2 - b^2 = 0      NB to get everything = 0 with a quadratic eqn. 

(x + b)(x - b) = 0

x + b = 0 Or x - b = 0

x = -b Or x = b

Term I've never heard of that term. Thanks for that. 

[TheoG]

I somma call it a 4th order polynomial ... 

 

Must say up untill 1st semester 3rd year maths, I never heard the term "quartic", maybe I was not in class that day  .

 

Edit:  Googled it, I definitely missed that lesson 

[TheoG]

1.

(5x^4)/3 - 20 = 0

5x^4 - 60 = 0

Divide everything by 5 and factorise

 

x^4 - 12 = 0

(x^2 - sqt12)(x^2 + sqtt12) = 0

x^2 - sqt12 = 0. OR x^2 + sqt12 = 0

Now you must factorise  the x^2 - sqt12 = 0 again

 

This is difficult to write on computer but you will essentially get x = +//12 and x = - //12

 

2. 

(3x)^2 +36^1/2 = 100

9x^2 + 6 = 100

9x^2 - 94 = 0

Now factorise

 

(3x - sqt94)(3x + sqt94) = 0

3x - sqt94 = 0 OR 3x + sqtt 94 = 0

x = (sqt93)/3. Or x = - (sqt94)/3

 

On the first one, because there is no x^3 term, there is only 2 roots (solutions), if it did have a x^3 term, can not be solved analytically but only by a numerical method such as Newton Rapson, but sorry, thats making it way more complicated than required ...

[greys]

that's nice. I study math in college but it's not so easy task for me. Thanks to the cheap essay writer service I got good grades and saved a lot of time. These guys helped me a lot.

[Slowbee]

https://www.sciencealert.com/math-genius-finally-discovers-easy-way-to-solve-quadratic-equations-after-4-000-years

 

#thefutureishere

[TheoG]

https://www.sciencealert.com/math-genius-finally-discovers-easy-way-to-solve-quadratic-equations-after-4-000-years

 

#thefutureishere

 

Very very cool, I love it !  That makes it very simple to solve quadratic equations that is not straight forward to factor.

[Slowbee]

Very very cool, I love it !  That makes it very simple to solve quadratic equations that is not straight forward to factor.

I am sure changing the school curriculum is going to be a problem - where "marks" are awarded for both steps and answer.

[TheoG]

True.  Not a problem for me though, I will just USE the method ..  

[TheoG]

Its important to note that for this new method "a = 1" is a must, if it is not, you need to divide the equation by "a" to make it "1".

 

With the old quadratic equations, matters not, "a" can be anything.

[rudi-h]

You're trying to do too much in one step, that means you'll always make mistakes.

 

Paper and ink is cheap, write down a single step at a time.  e.g. If you have the square root of 36 in one line, repeat the whole line and just change that single term to 6.

 

Nobody gets marks for writing down fewer steps, you only get marks for the right answer

[TheoG]

You're trying to do too much in one step, that means you'll always make mistakes.

 

Paper and ink is cheap, write down a single step at a time.  e.g. If you have the square root of 36 in one line, repeat the whole line and just change that single term to 6.

 

Nobody gets marks for writing down fewer steps, you only get marks for the right answer

 

I think the OP was sorted long ago ...  

[rudi-h]

Maybe sorted with that one problem, but the OP will struggle as much with any other problem that might be thrown his way.

 

My advice was offered as a general pointer applicable to any future problem in any domain that the OP might face...  He did ask for direction and guidance after all

 

I think the OP was sorted long ago ...  

[Long Wheel Base]

Just stumble upon this thread now. Geez when do the kids start with this at school? My oldest is in Grade R this year so I suspect(hope) I have a while before I come here asking for help.