How do you solve #sqrt(5x^2+11)=x+5# and identify any restrictions?
1 Answer
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
#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
#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
Check both of these by plugging them into the original equation:
Checking
#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
Checking
#sqrt(5(-1)^2+11)=-1+5#
#sqrt(5+11)=4#
#sqrt16=4#
Which is true as well, so our solutions are
We can skip the process of going back and checking answers by noting that since we have
Note that since