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).
@WahranRai
5 ай бұрын
8:11 2 numbers having product P = ab and the sum S = - (a+b)
@ScienceWorldz
5 ай бұрын
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
@CriticSimon
5 ай бұрын
It's the same thing :)
@yoav613
5 ай бұрын
Nice! It's also worth to notice that if a+b=0 (a=-b) then x can be any number not equal to 0.
@Khirvakka
5 ай бұрын
This is my 10 class math
@ryanrahuelvalentine2879
5 ай бұрын
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.
@giuseppemalaguti435
5 ай бұрын
x^2+(a+b)x+ab=0
@ManjulaMathew-wb3zn
5 ай бұрын
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.
@ManjulaMathew-wb3zn
5 ай бұрын
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
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