How do you solve #sqrt(a-2)+4=a# and check your solution?

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Answer 1 · 2017-01-17T03:58:13

#a=6#

We start with:

#sqrt(a-2)+4=a#

Let's bring the 4 over to the right side so that we can square the square root:

#sqrt(a-2)=a-4#

#(sqrt(a-2))^2=(a-4)^2#

#a-2=a^2-8a+16#

Now let's bring all the terms to one side and set it to 0:

#a^2-9a+18=0#

#(a-6)(a-3)=0#

therefore

#a=6, a=3#

And now let's check our solutions:

#sqrt(6-2)+4=6#

#sqrt(4)+4=6#

#2+4=6#

#6=6# #color(green)(Check!)#

~~~~~

#sqrt(3-2)+4=3#

#sqrt(1)+4=3#

#1+4=3#

#5!=3# #color(red)(No!)#

And so our final solution is #a=6#

We can also check this solution by graphing the left side and the right side:

graph{(y-sqrt(x-2)-4)(y-x)=0 [-4.79, 15.21, -1, 9]}