The original rational equation is symmetric in a, b, and x. So the multiplied out version at 4:01 can be written symmetrically also. 0 = (a+b)(x^2 +(a+b)x + ab) 0 = (a+b)(x+a)(x+b) So two of the variables have to be opposites (but not zero), and the third variable is free (but can't be zero).
First guy! Damn, you explain so nicely. By seeing your question solving videos, I also try them. By doing this I am able to solve miscellaneous math questions.
The original rational equation is symmetric in a, b, and x. So the multiplied out version at 4:01 can be written symmetrically also.
0 = (a+b)(x^2 +(a+b)x + ab)
0 = (a+b)(x+a)(x+b)
So two of the variables have to be opposites (but not zero), and the third variable is free (but can't be zero).
8:11 2 numbers having product P = ab and the sum S = - (a+b)
A simpler solution:
1/(a+b+x) = 1/a + 1/b + 1/x
Send 1/x to LHS,
1/(a+b+x) - 1/x = 1/a + 1/b
Taking LCM,
-(a+b)/(ax+bx+x²) = (a+b)/ab
Considering a+b≠0,
-1/(ax+bx+x²) = 1/ab
-ab = (a+b)x + x²
x² + (a+b)x + ab = 0
Therefore, solutions are x = -a, -b
It's the same thing :)
Nice! It's also worth to notice that if a+b=0 (a=-b) then x can be any number not equal to 0.
First guy! Damn, you explain so nicely. By seeing your question solving videos, I also try them. By doing this I am able to solve miscellaneous math questions.
This is my 10 class math
x^2+(a+b)x+ab=0
The equation can be rewritten as
x^2-(-a+(-b))x+(-a)(-b)=0 which is in the form of Vieta’s
So roots are -a and - b.
Even more simply if the sum of the roots is -(a+b) and the product of them is ab then the roots mus be
-a and -b
x=-a
x=-b
x≠0
Very ratsional