Math, asked by ayinampudisrinivasar, 1 year ago

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

Answers

Answered by mysticd
2

 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 }

•••♪

Similar questions