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Пікірлер: 16
@is77286 ай бұрын
7:03 it should be a^2 MINUS 2a + 1.
@ismettandirovic63523 ай бұрын
Sviđa mi se.
@user-bs8mi3qw7g2 ай бұрын
I think you do a terrific job explaining all the steps. Your methods may be a bit long, but they are very helpful to struggling math students. I taught high school math for 25 years and I always learn something from his videos. Please keep them coming. Dr. Kapadia 🌺
@rainerzufall425 ай бұрын
You were kind of lucky (because of the construction of the question), that the sum (a-1)² + (b-1)² + (c-1)² + (d-1)² was zero (0) and not some positive number! That's the only occasion, that you can follow, that all these squares are zero. For negative, there are no real solutions, for positive, you don't know how to continue. For zero, you get what you want. So what about sqrt(x-1)+sqrt(2x-3)+sqrt(3x-5)+sqrt(4x-7) = 5x - 6.005 ? Are you helpless now? The sum of squares is now 0.01 ! You'll get infinite many solutions now, e.g. a=1.1 or a=0.9 with b=c=d=1 or b=1.1 or b=0.9 with a=c=d=1 or ... and everything in between!
@rainerzufall425 ай бұрын
7:03 MINUS! a² - 2a + ...
@God-gi9iu6 ай бұрын
Wow
@user-dv8gv3hu4t6 ай бұрын
Solution by insight 1+1+1+1=10-6 x=2
@rvqx6 ай бұрын
Yes, this much easier to find than the given solution.
@bhabenbordoli19813 ай бұрын
P/ mention the theory that if sum of term s are zero then rach term is equel to zero.