Было объяснение! Оно для тех, кто начал изучать комплексные числа! И не забывайте, что четные степени дают обязательно положительные и отрицательные корни. Комментарии диванных математиков умиляют. Автору огромное спасибо. С удовольствием смотрю твои ролики.
Actually it does because the four fourth roots of 16 are 2, -2, 2i, and -2i. Thus, just add 3 to each of these to find all four solutions in just 2 steps.
In triangle ABC, which is acute and not isoceles There are points D, E, and F, which are the feet of the elevation lines that descends from vertices A, B, C A circle with diameter AD cuts DE, DF in sequence of M for DE, N for DF Choose points P, Q on AB (P), AC (Q) in such a way that NP is perpendicular to AB, MQ is perpendicular to AC (I) is the circumcircle tangent to the center of triangle APQ a) prove that (I) touches EF b) T is the point at which (I) touches EF, DT cuts MN at K. Let L be the symmetry point of A through MN. Prove that the circumcircle of triangle DKL goes through the intersection of EF and MN.
Why do you use different methods when removing the fourth power and when removing the square power? If we use the a^2-b^2 expansion again the next line would just be ((x-3)^2+2^2)((x-3)^2-2^2)=(x-3+2i)(x-3-2i)(x-3+2)(x-3-2)=0.
Было объяснение! Оно для тех, кто начал изучать комплексные числа! И не забывайте, что четные степени дают обязательно положительные и отрицательные корни. Комментарии диванных математиков умиляют. Автору огромное спасибо. С удовольствием смотрю твои ролики.
Why not just make 16 into plus or minus 2^4 so that x-3=2 and x-3=-2?
I guess that doesn't get the complex solutions.
yeah I thaught the same thing... with the 4th root of 16 right?
Actually it does because the four fourth roots of 16 are 2, -2, 2i, and -2i. Thus, just add 3 to each of these to find all four solutions in just 2 steps.
(x-3)quad = 16 -> square root((x-3)quad) = square root(16) -> (x-3)squared = 4 -> x-3 = square root (4) -> (x-3) = 2 -> (x-3)+3 = 2+3 -> x = 5. Mentally about 7 seconds but about a 1.5 minutes to type.
Why write so many things when you can find the forth root on both sides
You make things look hard for those learning mathematics
It’s because he’s explaining it for the math Olympiad and they are not allowed to use calculators so they cant find the fourth root without one
In triangle ABC, which is acute and not isoceles
There are points D, E, and F, which are the feet of the elevation lines that descends from vertices A, B, C
A circle with diameter AD cuts DE, DF in sequence of M for DE, N for DF
Choose points P, Q on AB (P), AC (Q) in such a way that NP is perpendicular to AB, MQ is perpendicular to AC
(I) is the circumcircle tangent to the center of triangle APQ
a) prove that (I) touches EF
b) T is the point at which (I) touches EF, DT cuts MN at K. Let L be the symmetry point of A through MN. Prove that the circumcircle of triangle DKL goes through the intersection of EF and MN.
Solution
(X-3)^4 = 16
(X-3) ^4= (-2) ^4 or 2^4
Taking 4th root both sides
X-3 = -2 or 2
X = -2+3 or 2+3
X= 1 or 5
0:13 Yakushin!
Why do you use different methods when removing the fourth power and when removing the square power?
If we use the a^2-b^2 expansion again the next line would just be
((x-3)^2+2^2)((x-3)^2-2^2)=(x-3+2i)(x-3-2i)(x-3+2)(x-3-2)=0.
Unnötig kompliziert gelöst. Einfach die 4. Wurzel ziehen und es muss nur noch x-3=2 gelöst werden. So kommt einfach x=5 raus.
X=5
And other 3 roots?
All roots 16^(1/4) are +/-2 and +/-2i. Shifting it by +3 due to subtraction gave me all solutions immediately. 🙂
Creative resolution! I just made (x-3)²= +-4 and found the 4 roots.
Your way work you just have to continue by setting them both to 0 which is what he did.
The hardest possible way
Bhai ye easy question tha Maine to ise dekhte hi solve kiya
Merci ..
Parfaite.demonstation👍🌹🇩🇿
Good
Как он любит все усложнять . Вопрос .ЗАЧЕМ !!!!
🙏
Note gud honestly tho
My friend your answer is wrong , the answer is 5 , if you take 5 and substitute for X the answer will work out 16
X=1 or 5
X-3=+-2. X=-1 and 5 it is very easy solutions
x=5