Question

root of3 (x-4)^2-5 (x-4)=12
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Answer 1

$Given \: Quadratic \:equation : \\3(x-4)^{2} -5(x-4)= 12$

$\implies 3(x-4)^{2} -5(x-4)-12=0$

/* Splitting the middle term, we get */

$\implies 3(x-4)^{2} -9(x-4)+ 4(x-4)-12=0$

$\implies 3(x-4)[(x-4)-3] + 4[(x-4)-3] = 0$

$\implies3(x-4)(x-7) + 4(x-7) = 0$

$\implies (x-7)[3(x-4)+4] = 0$

$\implies (x-7)(3x-12+4) = 0$

$\implies (x-7)(3x-8) = 0$

$\implies x-7 = 0 \:Or \: 3x-8= 0$

$\implies x = 7 \:Or \: x = \frac{8}{3}$

Therefore.,

$\green { 7 \: and \: \frac{8}{3}\:are \:roots \: of }\\\green { given \: equation }$

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