How do you solve the equation #4+sqrt(4t-4)=t#?

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Answer 1 · 2016-10-17T06:22:48

#t=2# or #t=10#

#4+ sqrt(4t-4)= t#

Subtract 4 from both sides.

#sqrt(4t-4)= t-4#

Square both sides.

#4t-4 = (t-4)^2#

Expand the right side.

#4t-4= (t-4)(t-4)#

#4t-4= t(t-4)-4(t-4)#

#4t-4=t^2-4t-4t+16# (since multiplying two negatives produces a positive)

Simplify the equation on the right.

#4t-4=t^2-8t+16#

Subtract #(4t-4)# from both sides.

#0=t^2-12t+20# or #t^2 -12t+20=0#

Factorise the equation.

#t^2-2t-10t+20=0#

#t(t-2)-10(t-2)=0#

#(t-2)(t-10)=0#

#:. t=2# or #t=10#