How do you solve #sqrt(6a-6)=a+1# and check your solution?

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Answer 1 · 2017-09-30T16:34:49

No solutions

or using imaginary numbers
#a=2+-sqrt(3)i#

To start we would square both sides to get rid of the square root.

#(sqrt(6a-6))^2=(a+1)^2#
#6a-6=(a+1)^2#

We can now expand the brackets.

#6a-6=a^2+2a+1#

Move everything to one side to get the equation equal to 0.

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

To solve this we will use the quadratic formula.

#x=(-b+-sqrt(b^2-4ac))/(2a)#
#a=(-(-4)+-sqrt((-4)^2-4(1)(7)))/(2(1))#
#a=(4+-sqrt(16-28))/2#
#a=(4+-sqrt(-12))/2#

However we are left with the square root of a negative, so there are no solutions.

Whereas if you are using imaginary number we can continue. If you are not using imaginary numbers stop here.

We can split #sqrt(-12)# into #sqrt(-1)sqrt(12)#

#a=(4+-sqrt(-12))/2#
#a=(4+-sqrt(-1)sqrt(12))/2#

Put #i# in as #sqrt(-1)# and simplify #sqrt12#.

#a=(4+-2sqrt(3)i)/2#
#a=2+-sqrt(3)i#