Question

solve this by quadratic equations​

Answer 1

❏ Answer:-

Given:-

$\bullet\normalsize\sf{\:\left(\dfrac{1}{x-2}+\dfrac{2}{x-1}\right)\:=\:\dfrac{6}{x}}$

$\mapsto\normalsize\sf{\:\left(\dfrac{(x-1)+2(x-2)}{(x-1)(x-2)}\right)\:=\:\dfrac{6}{x}}$

$\mapsto\normalsize\sf{\:\left(\dfrac{(x-1+2x-4)}{(x^2+2-x-2x)}\right)\:=\:\dfrac{6}{x}}$

$\mapsto\normalsize\sf{\:\dfrac{3x-5}{x^2-3x+2}\:=\:\dfrac{6}{x}}$

$\mapsto\normalsize\sf{\:\dfrac{x(3x-5)}{6(x^2-3x+2)}\:=\:1}$

$\mapsto\normalsize\sf{\:\dfrac{(3x^2-5x)}{(6x^2-18x+12)}\:=\:1}$

$\mapsto\normalsize\sf{\:(3x^2-5x)\:=\:(6x^2-18x+12)}$

$\mapsto\normalsize\sf{\:(6x^2-18x+12-3x^2+5x)\:=\:0}$

$\mapsto\normalsize\sf{\:(3x^2-13x+12)\:=\:0}$

$\mapsto\normalsize\sf{\:(3x^2-9x-4x+12)\:=\:0}$

$\mapsto\normalsize\sf{\:3x(x-3)-4(x-3)\:=\:0}$

$\mapsto\normalsize\sf{\:(x-3)(3x-4)\:=\:0}$

$\mapsto\normalsize\sf{\:(x-3)\:=\:0\:\:\:Or\:\:(3x-4)\:=\:0}$

$\mapsto\normalsize\pink{\sf{\:x\:=\:3\:\:\:\:Or\:\:\:x\:=\dfrac{4}{3}}}$

Answer 2

Answer:

∙(x−21+x−12)=x6

\mapsto\normalsize\sf{\:\left(\dfrac{(x-1)+2(x-2)}{(x-1)(x-2)}\right)\:=\:\dfrac{6}{x}}↦((x−1)(x−2)(x−1)+2(x−2))=x6

\mapsto\normalsize\sf{\:\left(\dfrac{(x-1+2x-4)}{(x^2+2-x-2x)}\right)\:=\:\dfrac{6}{x}}↦((x2+2−x−2x)(x−1+2x−4))=x6

\mapsto\normalsize\sf{\:\dfrac{3x-5}{x^2-3x+2}\:=\:\dfrac{6}{x}}↦x2−3x+23x−5=x6

\mapsto\normalsize\sf{\:\dfrac{x(3x-5)}{6(x^2-3x+2)}\:=\:1}↦6(x2−3x+2)x(3x−5)=1

\mapsto\normalsize\sf{\:\dfrac{(3x^2-5x)}{(6x^2-18x+12)}\:=\:1}↦(6x2−18x+12)(3x2−5x)=1

\mapsto\normalsize\sf{\:(3x^2-5x)\:=\:(6x^2-18x+12)}↦(3x2−5x)=(6x2−18x+12)

\mapsto\normalsize\sf{\:(6x^2-18x+12-3x^2+5x)\:=\:0}↦(6x2−18x+12−3x2+5x)=0

\mapsto\normalsize\sf{\:(3x^2-13x+12)\:=\:0}↦(3x2−13x+12)=0

\mapsto\normalsize\sf{\:(3x^2-9x-4x+12)\:=\:0}↦(3x2−9x−4x+12)=0

\mapsto\normalsize\sf{\:3x(x-3)-4(x-3)\:=\:0}↦3x(x−3)−4(x−3)=0

\mapsto\normalsize\sf{\:(x-3)(3x-4)\:=\:0}↦(x−3)(3x−4)=0

\mapsto\normalsize\sf{\:(x-3)\:=\:0\:\:\:Or\:\:(3x-4)\:=\:0}↦(x−3)=0Or(3x−4)=0

\mapsto\normalsize\pink{\sf{\:x\:=\:3\:\:\:\:Or\:\:\:x\:=\dfrac{4}{3}}}↦x=3Orx=34