Question

3(x-3)(x+4)+3(x-2)(x-4)=19(x-4)(x-3)​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{3(x-3)(x+4)+3(x-2)(x-4)=19(x-4)(x-3)}$

$\underline{\textsf{To find:}}$

$\textsf{Roots of the given polynomial equation}$

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{3(x-3)(x+4)+3(x-2)(x-4)=19(x-4)(x-3)}$

$\mathsf{Rearranging\;terms}$

$\mathsf{3(x-3)(x+4)=19(x-4)(x-3)-3(x-2)(x-4)}$

$\mathsf{3(x^2+x-12)=(x-4)[19(x-3)-3(x-2)]}$

$\mathsf{3x^2+3x-36=(x-4)[19x-57-3x+6]}$

$\mathsf{3x^2+3x-36=(x-4)[16x-51]}$

$\mathsf{3x^2+3x-36=16x^2-51x-64x+204}$

$\mathsf{3x^2+3x-36=16x^2-115x+204}$

$\mathsf{13x^2-118x+240=0}$

$\mathsf{13x^2-78x-40x+240=0}$

$\mathsf{13x(x-6)-40(x-6)=0}$

$\mathsf{(13x-40)(x-6)=0}$

$\implies\mathsf{13x-40=0\;\;(or)\;\;x-6=0}$

$\implies\mathsf{x=\dfrac{40}{13}\;\;(or)\;\;x=6}$

$\therefore\bf\mathsf{Solutions\;set\;is\;\{\dfrac{40}{13},6\}}$

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