How do you solve #(4a-1)(a-2)=7a-5#?

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Answer 1 · 2016-07-16T09:36:56

#a = 7/2 or 1/2#

Foil the bracket on the left hand side:

#(color(red)(4a) - color(blue)(1))(color(orange)(a) - color(purple)(2)) = 7a - 5#

#color(red)(4a)*color(orange)(a) - color(red)(4a)*color(purple)(2) - color(blue)(1)*color(orange)(a) + color(blue)(1)*color(purple)(2) = 7a - 5#

#4a^2 - 8a - a + 2 = 7a - 5#

#4a^2 - 16a + 7 = 0#

#a = (16+-sqrt(16^2 - 4(4)(7)))/(2(4))#

#a = (16+-sqrt(16(16-7)))/8#

#a = (16+-12)/8 = 28/8 or 4/8#

#therefore a = 7/2 or 1/2#