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Question

Solve the following quadratic equation by factorization.
axa+bxb=2cxc

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Solution

fraction numerator a left parenthesis x minus b right parenthesis plus b left parenthesis x minus a right parenthesis over denominator left parenthesis x minus b right parenthesis left parenthesis x minus a right parenthesis end fraction equals fraction numerator 2 c over denominator x minus c end fraction fraction numerator a x minus a b plus b x minus a b over denominator left parenthesis x squared minus b x minus a x plus a b right parenthesis end fraction equals fraction numerator 2 c over denominator x minus c end fraction

= (xc)(ax2ab+bx)=2c(x2bxax+ab)

=(a+b)x22abx(a+b)cx+2abc=2cx22c(a+b)x+2abc
=(a+b2c)x22abx+acx+bcx=0
=x((a+b2c)x2ab+ac+bc)=0

x=2abacbca+b2c


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