@@woodylipinski9063Not so much a “guess,” but I mentally checked through the perfect squares, then 49 solved the puzzle. The procedure listed above in this comment thread is also the best way to do it by running through the equation.
There is a far better way to solve this: Using the fact that a^2 - b^2 = (a-b) *(a+b) let root(x) = a and root(x-40)=b So a+b= 10. a^2-b^2 = x - (x-40)=40 a-b = (a^2 - b^2) / (a+b) = 40/10= 4 Now add (a+b)+ (a-b)= 2a = 10+ 4 = 14 -> a=7 x = 7^2 =49 I would like to claim this as my own method but it’s taken from the famous Hall and Knight higher algebra book. A key advantage of the method is that it does not introduce false solutions which can appear when you square equations. Btw I was able to solve this mentally and was almost able to type the solution out in the time taken on the video.
Another way to say the same solution is just that you are rationalizing the numerator... Rationalizing the numerator gives: 10= root(x) + root(x-40) = 40 / (root(x) - root(x-40)) Thus root(x) - root(x-40) = 4. Then adding equations gives 2root(x) = 14 and x=49, as you said. I like to word things this way when teaching, because it feels less ad hoc since it connects back to earlier methods in the course, and it also foreshadows the trick to taking the derivative of root(x) in calculus.
Letting y= sqrt(x) implies y + sqrt(y^2-40) = 10 hence sqrt(y^2-40) = 10-y y^2-40 = (y-10)^2 [Equation A] on the left side, it’s convenient to complete the square (y-10)^2 (i.e. the same we have on the right side), by adding and subtracting some stuff: y^2 - 40 = y ^2 - 2*10y + 10^2 + 2*10y - 10^2 - 40 = (rearranging) = (y^2-2*10y+10^2) +2*10y -10^2 -40 = (y-10)^2 +20y -140 so, [Equation A] becomes: (y-10)^2 +20y -140 = (y-10)^2 and, simplifying: 20y -140 = 0, then y= 7 hence x = y^2 = 49
x^1/2+(x_40)^1/2=10 Solution since (10) in the right is integer number so (x_40)^1/2 must be integer and the nearest number is 49 . (49)^1/2+ (49 _40)^2/2=7+(9)^1/2 = 7+3=10 answer
Rather clumsy solution. More importantly, you must be careful when you square an identity. X=4 is not equivalent to X^2=4^2 which has 2 roots X=4 and X=-4. In your stuff you are lucky because the square terms cancel out, which ensures equivalence by linearity. Failing to mention this is misleading.
Bring √x to the RHS and then square both sides :
x - 40 = 100 - 20√x + x
-40 = 100 - 20√x
20√x = 140
√x = 7
x = 49
or guess 7 + 3 = √49 +√x-40
@@woodylipinski9063Not so much a “guess,” but I mentally checked through the perfect squares, then 49 solved the puzzle. The procedure listed above in this comment thread is also the best way to do it by running through the equation.
Visually at a glance the answer should be 49
There is a far better way to solve this:
Using the fact that a^2 - b^2 = (a-b) *(a+b)
let root(x) = a and root(x-40)=b
So a+b= 10.
a^2-b^2 = x - (x-40)=40
a-b = (a^2 - b^2) / (a+b) = 40/10= 4
Now add (a+b)+ (a-b)= 2a = 10+ 4 = 14
-> a=7
x = 7^2 =49
I would like to claim this as my own method but it’s taken from the famous Hall and Knight higher algebra book. A key advantage of the method is that it does not introduce false solutions which can appear when you square equations.
Btw I was able to solve this mentally and was almost able to type the solution out in the time taken on the video.
Diferença de dois quadrados é (a-B)*(a+b)
Another way to say the same solution is just that you are rationalizing the numerator... Rationalizing the numerator gives:
10= root(x) + root(x-40) = 40 / (root(x) - root(x-40))
Thus root(x) - root(x-40) = 4.
Then adding equations gives 2root(x) = 14 and x=49, as you said.
I like to word things this way when teaching, because it feels less ad hoc since it connects back to earlier methods in the course, and it also foreshadows the trick to taking the derivative of root(x) in calculus.
This equation is boring, you notice x=49 satisfy, then if x>49,LHS>10. If 40
Easy way is to make a substition. Either let X=M² or X=M²+40. That will eliminate one of the square roots and the algebra becomes much simpler.
Letting y= sqrt(x)
implies
y + sqrt(y^2-40) = 10
hence
sqrt(y^2-40) = 10-y
y^2-40 = (y-10)^2 [Equation A]
on the left side, it’s convenient to complete the square (y-10)^2 (i.e. the same we have on the right side), by adding and subtracting some stuff:
y^2 - 40 = y ^2 - 2*10y + 10^2 + 2*10y - 10^2 - 40 = (rearranging) = (y^2-2*10y+10^2) +2*10y -10^2 -40 = (y-10)^2 +20y -140
so, [Equation A] becomes:
(y-10)^2 +20y -140 = (y-10)^2
and, simplifying:
20y -140 = 0,
then
y= 7
hence
x = y^2 = 49
❤❤Gracias, aprendo mucho❤❤❤
Start taking sqrt(X) to the RHS so you get sqrt(X - 40) = 10 - sqrt(x) then square each side. It then becomes a simple equation to solve!
Got it before I watched the vid. Good practice😎
x^1/2+(x_40)^1/2=10
Solution
since (10) in the right is integer number so (x_40)^1/2 must be integer and the nearest number is 49 .
(49)^1/2+
(49 _40)^2/2=7+(9)^1/2 =
7+3=10 answer
Hello, this was super easy
I solved it under 5 seconds without the need of writing.
teach me
@@Kayky_Nunes He probably guessed or tried it. x must be a square number larger than 40 and coincidentally the first one fits: 49
The function sqr(x)+sqr(x-40) is strictly rising so the equation cannot have more than one sulition, whis is 49
Bring sqrt(x-40) to the RHS and then square both sides.
Then RHS will be quite complicated
Bring sqrt(x) to RHS instead
@shantanudhiman8194 It seems useful to get the square root of x to the right side and then square it.
49
Grazie
No more interested in X, I will find next instead!!!😄😁😆😅🤣
２回目の２乗するとき、
両辺０以上よりXは70以下。
X=49より条件を満たす。
Let y=x-20, then sqrt(y+20)+sqrt(y-20)=10 would be much easier to solver.
A no. Whoes square root in near 10 ..64 or 49..the moment we subtract 9 ..there it's 49
Решение неверное примерно со второй строчки. Дальше тоже куча ошибок, но я про это уже молчу
Buen ejercicio, pero puedes abreviarlo
10=7+3=√x+√(x-40),x-40>0,√x=7,x=49>40,
√(49-40)=3にて、x=49は適す．∴x=49
x is the first 2 letters of the equation
Решение не является правомерным пока не установлены ограничения
I'm 72years learning Maths as a hobby. Why is the answer not 70?
Le me legend : giving the answer just by observation even before starting the video 🤣🤣.
Madam plz mujy ap ye btay ke ap thumbnail kaisy bnati hai
Why do you need to calculate this? Anyone can see right away x = 49
X=49
no restrictions no right answer , x=49 by luck
You are dragging it. You can literally explain. But make it fast. 🤷🏻🤷🏻
Too slow. Be in cadence with the music at least.
Too long from the beginning you could pass square x to the second member is faster
Х=49
Are you blind. X is inside the square root sign😂
Rather clumsy solution.
More importantly, you must be careful when you square an identity. X=4 is not equivalent to X^2=4^2 which has 2 roots X=4 and X=-4. In your stuff you are lucky because the square terms cancel out, which ensures equivalence by linearity. Failing to mention this is misleading.
Method too long and elementary
Сам(а) то научись сначала делать по быстрее и без ошибок
49
49
49