How do you solve #sqrt(5x^2+11)=x+5# and identify any restrictions?

1 Answer
Mar 19, 2017

Answer:

#x=-1# and #x=7/2#. This equation has no restrictions since #5x^2+11>=0#.

Explanation:

Begin by squaring both sides of the equation to eliminate the square root. Recall that in squaring the right hand side, we can't square each term individually--we have to square the entire right hand side.

#(sqrt(5x^2+11))^2=(x+5)^2#

On the left, the #sqrt# and #""^2# cancel. On the right, expand the square by FOILing:

#5x^2+11=(x+5)(x+5)=x^2+5x+5x+25#

#5x^2+11=x^2+10x+25#

Putting all the terms on the same side:

#4x^2-10x-14=0#

Divide all terms by #2#:

#2x^2-5x-7=0#

Which we can factor by splitting the middle term:

#2x^2+2x-7x-7=0#

#2x(x+1)-7(x+1)=0#

#(2x-7)(x+1)=0#

Which give #x=7/2# and #x=-1#.

Check both of these by plugging them into the original equation:

Checking #x=7/2#:

#sqrt(5(7/2)^2+11)=7/2+5#

#sqrt(5(49/4)+11)=7/2+10/2#

#sqrt(245/4+44/4)=7/2+10/2#

#sqrt(289/4)=17/2#

Which is true! So #x=7/2# is a valid solution.

Checking #x=-1#:

#sqrt(5(-1)^2+11)=-1+5#

#sqrt(5+11)=4#

#sqrt16=4#

Which is true as well, so our solutions are #x=-1# and #x=7/2#.

We can skip the process of going back and checking answers by noting that since we have #sqrt(5x^2+11)# in the original equation, we can't have values of #x# where #5x^2+11<0#, since we can't take the square root of a negative value.

Note that since #5x^2# and #11# are both always positive, #5x^2+11# will also always be positive. Thus, there is never a time when it's negative and by the same coin there is never a time when #sqrt(5x^2+11)# won't be defined. Thus there are no restrictions on this equation.