How do you solve #sqrt( 2x+1)=x-3#?

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Answer 1 · 2015-08-05T09:43:18

This equation has 2 solutions: #x_1=4-2sqrt(2)# and #x_2=4+2sqrt(2)#

In the equation there is a square root, so first you have to calculate the domain.

In this case the experssion under root sign is #2x+1# so you have the domain: #2x+1>=0# or #x>=-1/2#

Now we can calculate the solutions.
We start with raising both sides to the second power

#2x+1=(x-3)^2#

#2x+1=x^2-6x+9#

#-x^2+8x-8=0#

Now we can solve the quadratic equation:

#Delta = 8^2-4*(-1)*(-8)#

#Delta=64-32=32#

#sqrt(Delta)=sqrt(32)=4sqrt(2)#

#x_1=(-8-4sqrt(2))/(2*(-1))#

#x_1=4+2sqrt(2)#

#x_2=(-8+4sqrt(2))/(2*(-1))#

#x_2=4-2sqrt(2)#

Now we have to check if any of calculated solutions are in the domain.

#x_1>4# so it surely is greater than #-1/2#

#x_2=4-2sqrt(2) ~~4-2*1,41~~4-2,82~~1,18>=-1/2#

We checked that both solutions are in the domain, so we can write the answer:

This equation has 2 solutions: #x_1=4-2sqrt(2)# and #x_2=4+2sqrt(2)#