(x+2)(x+3)(x+4)(x+5)=(x-2)...(x-5) 4th power terms are canceled on both sides and cubic term is (2+3+4+5)x^3 and -14 on the right so 28 is the cubic coefficient. x^2 terms are also the same on the right and left side so they cancel each other. likewise constant terms. and 2(2*3*4*5)(1/2+1/3+1/4+1/5)= 120+80+60+48=308 is the linear term. Divide by 28 and you have x^3+11x=0 ; x=0 or x^2+11=0
В уме эта задача решается так: Чтобы получилось 1, надо чтобы верняя часть была равна нижней. Они и с x равны, но остается знак минуса снизу. Но есть правило умножения. Любое число кроме 0 меняет равенство верхнего и нижнего. Вот и все решение. Знание правил арифметики и 1 минута в уме. Так учат считать в России 🎉
Nice job. I love seeing alternative methods to what some see as run-of-the-mill algebra problems. Thanks for sharing, and I will share this with my students.
@@KrytenKoroIt’s obvious that there can be no other real roots, because there is no other way to balance the numerator and denominator using real numbers. All other roots must be imaginary, and much have real part = 0
The even power terms, including the constant, can immediately be discarded, as they appear with same sign on both side. This leaves you with a simple equation of the form x(ax^2 + b) = 0, which is readily solvable.
Sometimes, the journey itself holds more value than the destination Solving obviously simple math in a long and exhaustive way can help build discipline and improve cognitive abilities as the brain is given a workout. However, one must also remember that simplicity is better than complexity
This title and these comments are why those who struggle with math are frustrated. My issue is context. I understand that this is the basics of calculating other math, but calculating random numbers is pointless. How does solving this help us understand other calculations?
😆 it is certainly a good suggestion, and it is another way to solve it, in this case we use variables to solve it step by step. Thanks for commenting my friend!
Zakonisht ne algjeber mesojme per grupe tipike dhe jane krijuar formula ndihmese . Kurse ky rast nuk eshte i tille, besoj une , por e krijojme si interesant.
Maybe we could let a=X^2+7X, b=X^2-7X and thus a+b=2X^2 , a-b=14X. With above equations, we will have (a+10)(a+12)=(b+10)(b+12) a^2-+22a=b^2+22b a^2-b^2=22(b-a) 2X^2*14X=-22*14X 14X(2X^2+22)=0 So X=0 is the only real root.
True, but if you remember the original problem, it specified x was a real number. We will not be dealing with the complex world today. However, if this had not been specified earlier, +/- i * sqrt(11) would be valid solutions.
You are totally wrong! 2x^2=-22 ; i,e x^2=-11 ; i,e x=√11i. or x=-√11i where i=√(-1). x has three values. One value is real (x=0) and other two values (√11i & -√11i) are complex numbers. There is no mention in that problem that x is only real. The problem should have mentioned that x is only real.
(x+2)(x+3)(x+4)(x+5) ÷ (x-2)(x-3)(x-4)(x-5) = (x⁴ + 14x³ + 71x² + 154x + 120) ÷ (x⁴ - 14x³ + 71x² - 154x + 120) Since the numerators are the same except for the signs, we can simplify the division: = (-1) The result is -1.
Also: => (x+2)(x+3)(x+4)(x+5)=(x-2)(x-3)(x-4)(x-5) => (x+2)(x+3)(x+4)(x+5)= (-1)⁴[(-x)+2][(-x)+3][(-x)+4][(-x)+5] => func.I (x+2)(x+3)(x+4)(x+5)= (a+2)(a+3)(a+4)(a+5) func.II; a= -x Thinking: func.I looks like func.II and a = -x Then, for that two things to be true, we need a number that: a = -x AND a = x => x- = x => 2x = 0 => x = 0 [ this root can be easy😀 ] For the other two roots we need a diferent approach 😅
I said we has 3 roots because 1) a= -x 2) func.I and func.II looks like the same 3) The functions are 4power So, functions have this desing: ax⁴+bx³+cx²+dx+e The sencond and the fourth terms not canceled each other, but the other terms go away. Then, we have: bx³+dx= -bx³-dx => 2(bx³+dx)=0 => bx³+dx=0 => x(bx²+d)=0 but, we know that x=0 can be true. So, cosidering the other term: => bx²+d=0 => x²= - (d/b) => x = ±i√(d/b); { with d € R and b € R }
Also: =>(x+2)(x+3)(x+4)(x+5)=(x-2)(x-3)(x-4)(x-5) => (x+2)(x+3)(x+4)(x+5) = (-1)⁴[(-x)+2][(-x)+3][(-x)+4][(-x)+5] =>(x+2)(x+3)(x+4)(x+5)=(a+2)(a+3)(a+4)(a+5); a = -x => x⁴+(2+3+4+5)x³+(2*3+2*4+2*5+ 3*4+3*5+4*5)x²+(2*3*4+2*3*5+ 2*4*5+3*4*5)x+(2*3*4*5)= the same for a => x⁴+14x³+71x²+154x+120 = a⁴+14a³+71a²+154a+120 =>x⁴+14x³+71x²+154x = x⁴-14x³+71x²-154x; {remember that a=-x and (-1)ⁿ=1, when n is pair} => 14x³+154x= -(14x³+154x) => 2(14x³+154x) = 0 => 14x³+154x=0 => 14(x³+11) = 0 => x³+11= 0 => x(x²+11) = 0 => x = 0 or x²+11 = 0 => x = 0 or x=i√11 or x=-i√11
Very elegant resolution
Thanks my friend! 👍🏻
(x+2)(x+3)(x+4)(x+5)=(x-2)...(x-5) 4th power terms are canceled on both sides and cubic term is (2+3+4+5)x^3 and -14 on the right so 28 is the cubic coefficient.
x^2 terms are also the same on the right and left side so they cancel each other. likewise constant terms.
and 2(2*3*4*5)(1/2+1/3+1/4+1/5)= 120+80+60+48=308 is the linear term. Divide by 28 and you have x^3+11x=0 ; x=0 or x^2+11=0
Excelente questão!!
В уме эта задача решается так:
Чтобы получилось 1, надо чтобы верняя часть была равна нижней. Они и с x равны, но остается знак минуса снизу. Но есть правило умножения. Любое число кроме 0 меняет равенство верхнего и нижнего. Вот и все решение. Знание правил арифметики и 1 минута в уме. Так учат считать в России 🎉
Тоже в уме посчитала)
I think too
А почему у него 7х=-7х?
Nice job. I love seeing alternative methods to what some see as run-of-the-mill algebra problems. Thanks for sharing, and I will share this with my students.
Thank you so much! 👍🏻 Glad it was helpful! Are you a teacher of maths?
If you know how to use a number line, you can solve this math problem in one minute.
How?
Solution by insight
2×3×4×5=(-2)×(-3)×
(-4)×(-5)
x=0
Gotta make sure you find all roots
@@KrytenKoroIt’s obvious that there can be no other real roots, because there is no other way to balance the numerator and denominator using real numbers. All other roots must be imaginary, and much have real part = 0
@@geoffstrickler This is an intuitive argument which has to be proven in formal terms. Nothing is obvious in math.
@ It is, and it’s easy to prove.
@@geoffstrickler Of course. By solving this problem.
Beautiful way..
Thank you so much 😁
Great
Excelent vídeo
Obrigado meu amigo! 👍🏻
Good demonstration
Thank you so much! Thanks for watching 👍🏻
The even power terms, including the constant, can immediately be discarded, as they appear with same sign on both side.
This leaves you with a simple equation of the form x(ax^2 + b) = 0, which is readily solvable.
The quadratic formula was pretty unnecessary, since you only had one x term, but good show!
Thanks for watching my friend! 😄
Without any effort X is shouting i'm equal to Zero 🤨
직관으로 x=0
Multiplying both sides by the denominator, adding 1 to both sides, and taking the square roots of them, it results: (x+2)*(x+5)+1=(x-2)*(x-5)+1.
Excellent! Thanks for watching 👍🏻
Very clever solution, although you didn't need to use the quadratic formula to solve Case 2
Thanks for watching my friend 👍🏻
Sometimes, the journey itself holds more value than the destination Solving obviously simple math in a long and exhaustive way can help build discipline and improve cognitive abilities as the brain is given a workout. However, one must also remember that simplicity is better than complexity
X = 0. Easy.
Is there other simpler way i can solve this without delta?
b non-zero has to be
Elegant - but fails to reduce solution time!
This title and these comments are why those who struggle with math are frustrated. My issue is context. I understand that this is the basics of calculating other math, but calculating random numbers is pointless. How does solving this help us understand other calculations?
Dude, this is boring. 2x^2=-22 so x=+-sqrt-22. End of story.
±√-11*
In your righteous indignation, you forgot to divide by 2.
It's true
Dude roasted himself
x=±sqrt(-11)
Yeah, x is large .. so the2,3,4 etc drop out. X4/x4 =1. End of story. I understand it’s nice to show more rigorous for students for practice. THX.
By inspection x = 0. Multiplying four negative numbers in the basement will give the same positive number upstairs. 120/120 = 1
Thanks for watching 👍🏻
It is easier to solve it directly without using a and b. Just do the multiplications with the order (x+2)(x+5).... and then simplify
Thanks for watching 👍🏻
@@MathmentorX19But you have no comment about the suggested METHOD? 🙄
😆 it is certainly a good suggestion, and it is another way to solve it, in this case we use variables to solve it step by step. Thanks for commenting my friend!
😂😂😂😂
😂😂😂😂
...×=0
I just simply multiply both side by a minus and there will be matching elements and cut them😁
That's great 👍🏻
Remarkable effort ❤
Thanks for watching ❤️
Zakonisht ne algjeber mesojme per grupe tipike dhe jane krijuar formula ndihmese . Kurse ky rast nuk eshte i tille, besoj une , por e krijojme si interesant.
😮
There is more four real solutions.
(-7±√5)÷2 and (7±√5)÷2.
Thanks for watching 👍🏻
Maybe we could let a=X^2+7X, b=X^2-7X and thus a+b=2X^2 , a-b=14X.
With above equations, we will have
(a+10)(a+12)=(b+10)(b+12)
a^2-+22a=b^2+22b
a^2-b^2=22(b-a)
2X^2*14X=-22*14X
14X(2X^2+22)=0
So X=0 is the only real root.
That's great! Thanks for watching 👍🏻
Eso salia al ojo xd
Si, pero también podríamos incluir soluciones no reales. Gracias por comentar. 👍🏻
let put x2 +7x+11 = a and x2 -7x+11 = b
2x^2 + 22 =0 . X^2 +11 =0.
Result is x equal to 0. I develop the whole expression
It's correct 👍🏻 thanks for commenting!
@ thank you for your great videos! Please continue 👏
X=0
use this
pochhammer(x+2,4)/pochhammer(x-5,4)=1
x1= 0 (real)
x2= -i sqrt(11) (complex)
x3= i sqrt(11) (complex)
Excellent! Thanks for watching 👍🏻
حلقة مفرغة كان من البداية اختصار والناتج 1
+√11i and -√11i are 2 imaginary solutions
Excellent! Thanks for watching 👍🏻
@MathmentorX19 😊
True, but if you remember the original problem, it specified x was a real number. We will not be dealing with the complex world today.
However, if this had not been specified earlier, +/- i * sqrt(11) would be valid solutions.
@stevendebettencourt7651 it's true 👍🏻
@@stevendebettencourt7651 yea thanks bro 😊
(x² + 7x + 10)(x² + 7x + 12) = (x² - 7x + 10)(x² - 7x + 12)
x² + 7x + 11 = u
x² - 7x + 11 = v
(u - 1)(u + 1) = (v - 1)(v + 1)
u² - 1 = v² - 1
u² = v² => u = ± v
u = v
x² + 7x + 11 = x² - 7x + 11
*x = 0*
u = -v
x² + 7x + 11 = - (x² - 7x + 11)
2x² + 22 = 0 => x² = -11
*x = ± i√11*
You are totally wrong! 2x^2=-22 ; i,e x^2=-11 ; i,e x=√11i. or x=-√11i where i=√(-1). x has three values. One value is real (x=0) and other two values (√11i & -√11i) are complex numbers. There is no mention in that problem that x is only real. The problem should have mentioned that x is only real.
It is mentioned in the starting given as, x ε R
Thầy giải máy móc quá, bài toàn này chỉ chuyển vế và lập luận đk của đẳng thức là lấy được nghiệm
Good! Thanks for watching 👍🏻
(x+2)(x+3)(x+4)(x+5) ÷ (x-2)(x-3)(x-4)(x-5) =
(x⁴ + 14x³ + 71x² + 154x + 120) ÷ (x⁴ - 14x³ + 71x² - 154x + 120)
Since the numerators are the same except for the signs, we can simplify the division:
= (-1)
The result is -1.
2x^2 + 22 = 0
x^2 + 11 = 0
x^2 = -11 no real solution.
Easier
Did not understand why you add +0x and reconvert to a more complex formula with a and b
Thanks for watching and commenting 👍🏻
Более тупого и затянутого решения я не видел. До здравствует советская школьная математика!!!
Hahahah 🤣 thanks for commenting my friend! 👍🏻
Also:
=> (x+2)(x+3)(x+4)(x+5)=(x-2)(x-3)(x-4)(x-5)
=> (x+2)(x+3)(x+4)(x+5)=
(-1)⁴[(-x)+2][(-x)+3][(-x)+4][(-x)+5]
=> func.I (x+2)(x+3)(x+4)(x+5)=
(a+2)(a+3)(a+4)(a+5) func.II; a= -x
Thinking:
func.I looks like func.II and a = -x
Then,
for that two things to be true, we need a number that:
a = -x AND a = x
=> x- = x
=> 2x = 0
=> x = 0 [ this root can be easy😀 ]
For the other two roots we need a diferent approach 😅
I said we has 3 roots because
1) a= -x
2) func.I and func.II looks like the same
3) The functions are 4power
So,
functions have this desing:
ax⁴+bx³+cx²+dx+e
The sencond and the fourth terms not canceled each other, but the other terms go away. Then, we have:
bx³+dx= -bx³-dx
=> 2(bx³+dx)=0
=> bx³+dx=0
=> x(bx²+d)=0
but, we know that x=0 can be true. So,
cosidering the other term:
=> bx²+d=0
=> x²= - (d/b)
=> x = ±i√(d/b);
{ with d € R and b € R }
x =/= { 2 , 3 , 4 , 5 }
t = x^2 + 7x + 10
t( t + 2 ) = ( t - 14x )( t - 14x + 2 )
28x ( 7x - t - 1 ) = 0
1) x = 0 o.k.
2) 7x - x^2 - 7x - 10 - 1 = 0
x^2 + 11 = x^2 - 11 i^2 = 0
x = +/- i sqrt( 11 )
no soluzioni reali
👍😁👋
Excellent! Thanks for watching 👍🏻
Also:
=>(x+2)(x+3)(x+4)(x+5)=(x-2)(x-3)(x-4)(x-5)
=> (x+2)(x+3)(x+4)(x+5) =
(-1)⁴[(-x)+2][(-x)+3][(-x)+4][(-x)+5]
=>(x+2)(x+3)(x+4)(x+5)=(a+2)(a+3)(a+4)(a+5); a = -x
=> x⁴+(2+3+4+5)x³+(2*3+2*4+2*5+ 3*4+3*5+4*5)x²+(2*3*4+2*3*5+ 2*4*5+3*4*5)x+(2*3*4*5)= the same for a
=> x⁴+14x³+71x²+154x+120 =
a⁴+14a³+71a²+154a+120
=>x⁴+14x³+71x²+154x =
x⁴-14x³+71x²-154x;
{remember that a=-x and (-1)ⁿ=1, when n is pair}
=> 14x³+154x= -(14x³+154x)
=> 2(14x³+154x) = 0
=> 14x³+154x=0
=> 14(x³+11) = 0
=> x³+11= 0
=> x(x²+11) = 0
=> x = 0 or x²+11 = 0
=> x = 0 or x=i√11 or x=-i√11