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x=Sqrt[27/(x-4)-20/(x-3)] x=1 x=2.5±0.5Sqrt[21]=(5±Sqrt[21])/2
Thanks for this one. Fourth order polynomial so 4 roots (Gauss - first fundamental theorem of algebra). (x-1)^2 is a factor. So x=1 is a repeated root.
It’s in my head.
Svolgo i calcoli..x^2=(7x-1)/(x-4)(x-3)...x^4-7x^3+12x^2-7x+1=0..(x-1)^2(x^2-5x+1)=0...
X1=1, X2=[5+(21)^(1/2)]/2, X3 =[5-(21)^(1/2)]/2
Need to check the answers for extraneous solutions when squaring both sides
Actually check that x>=0.
x= 1 , (5+V21)/2 , (5-V21)/2 ,
I got the same answers by solving the 4th degree equation: x^4-7x^3+12x^2-7x+1=0
X=1 is already root
B is NOT P !!! What is the matter with you speakers in these videos !! Can you not distinguish letters !!!???????
I'm working smart around the clock to abolish my pronunciation.
@@superacademy247 Good.
{27(x-3)-20(x-4)}/(x-3)(x-4) = (27x-81-20x+80)/(x-3)(x-4)=(7x-1)/(x^2-7x+12) ; i,e (7x-1)=x^2(x^2-7x+12) ; i,e 7x-1= x^4-7x^3+12x^2 ; i,e x^4-7x^3+12x^2-7x+1=0 ;i,e (x-1)(x^3-6x^2+6x-1)=0 ; (x-1)^2(x^2-5x+1)=0 ; i,e x=1 or x=1 or x= (5+√21)/2 or x= (5-√21)/2