Type a math problem and/or solve a math problem

Diophantine simultaneous equations are also cool.

Here's one... this problem concerns NOT finding the answer (it is trivially easy to solve just by trying a few different test values) but finding the METHOD by which this entire class of problems can reliably be solved.

Suppose you are asked to provide two different numbers, which add to give you 20 and multiply to give you 96.

As I said, it is trivially easy to solve this problem just by trying various whole-number factors of 96; and you will quickly find that the required two numbers are 12 and 8.

But can you find a METHOD which will allow you to solve every problem in this class, WITHOUT any guesswork or any testing of different numbers?

The method is below:

The conditions of the problem give you a pair of simultaneous equations:

x + y = 20
x * y = 96

Solving this pair is a bit awkward, since you have two different variables and the two equations use different operators (addition in one case, multiplication in the other). The trick is to reduce it to a pair of equations in a single variable, which is much more easily solved.

Every pair of numbers x and y can be represented as an average value n (exactly halfway between them) and a displacement d (representing how far the two "ends" are from the "middle"). So x and y can be substituted by:

x = n + d
y = n - d

Then our two simultaneous equations take on a new form:

(n + d) + (n - d) = 20
(n + d) * (n - d) = 96

The first equation reduces immediately:

(n + d) + (n - d) = 20

2n = 20
n = 10

And that gives us a value we can substitute into our second equation:

(n + d) * (n - d) = 96

(10 + d) * (10 - d) = 96
100 - d^2 = 96
100 = d^2 + 96
4 = d^2
d = +/- 2

So now we know that n = 10 and d = plus-or-minus 2. This allows us to calculate x and y directly:

x = 10 + 2
x = 12

y = 10 - 2
y = 8

And our method provides exact answers... no guesswork, no testing of different values.

x^2-20x+96=0 is the same by Viete coefficients formulas. If you don’t want to factor, use quadratic equation. No guess.

You can also solve that version of it by completing the square (although one might argue that using the General Quadratic IS completing the square).

You can also replace y by 20-x in the second equation to obtain the same.

Yeah, there are almost always multiple ways to approach any problem.

Completing the square with general coefficients a, b and c gives the quadratic formula.

Yes, that's exactly what I meant by my Post #23. Using the General Quadratic IS completing the square.

If you've scrolled this far maybe you've seen emu's answer, if not, then don't read the next sentence

"Every pair of numbers x and y can be represented as an average value n (exactly halfway between them) and a displacement d (representing how far the two 'ends' are from the "middle')"

Very cool, I've never seen a substitution like that.

You don’t get a formula if you complete the square with specific coefficients (numbers). A formula is a general solution for any set of coefficients (a, b, c).

This is like when we say the summit is at -b/(2a) and we add  and substract half the distance between the roots.

... and like I said, it enables you to solve a whole class of problems quickly and easily.

God's answer is 2.

x+y=2

x*y=1

x=-y+ 2

(-y+2)*y=1

-y^2+2y=1

y^2-2y+1=0

(y-1)^2=0

y-1=0

y=1

x=-1+2

x=-1

(-1,-1)

X / 20 + 5 * 5 = 45

X/20 + 25 = 45

X/20 = 20

X = 400

Cotx(secx^2- 1)

1 + 2 * 3 - 4 / 5

1 + 6 - 4 / 5

1 + 6 - 4/5 (4/5 being four fifths.)

7 - 4/5 

6 1/5 

1 + 2 * 3 - 4 / 5 = 6 1/5 (could also be written as 6.2

2

-1 + -1 ≠ 2

you missed -1+2 = 1

Try blueemu's method.

tan(x).

It would make 1 if you had a square for cot(x) too. Not defined for x=kπ/2, k ∈ ℤ.