How do you simplify #x + sqrt(x + 5) = 7#?

To simplify #x+sqrt(x+5)=7#, first transpose #x# to the RHS, i.e.

#sqrt(x+5)=7-x#

Hence, squaring the two sides, we get

#(x+5)=(7-x)^2# or #x+5=49-14x+x^2#

and transposing all terms to one side #x^2-15x+44=0#

and then splitting middle term in #-11x# and #-4x#, we have

#x^2-11x-4x+44=0# or #x(x-11)-4(x-11)=0# or

#(x-4)(x-11)=0#

i.e #x=4# or #x=11#

Note that #x=4# works when we take positive root of #sqrt(x+5)# and #x=11# works only when we take negative root of #sqrt(x+5)#

#sqrt(x + 5) = 7 - x#

#(sqrt(x + 5))^2 = (7 - x)^2#

#x + 5 = 49 - 14x + x^2#

#0 = x^2 - 14x - x + 49 - 5#

#0 = x^2 - 15x + 44#

Factor to solve. To factor a trinomial of the form #y = ax^2 + bx + c, a= 1#, we must find two numbers that multiply to c and that add to b. -11 and -4 are two numbers that work.

#0 = (x - 11)(x - 4)#

#x = 11 and 4#

We must always check our solutions in the original equation, since solutions known as extraneous solutions. These are solutions that occur when squaring the equation and that don't work inside the original equation. We you find a solution that doesn't work, you must always reject it. Checking our solutions from the previous equation I've found that #x = 11# is an extraneous solution. Thus, our solution set is #{x = 4}#.

Practice exercises:

1. Solve for x.

a). #sqrt(2x + 1) = 10 - x#

b). #sqrt(3x + 1) + 5 = sqrt(16x + 1)#

c). #sqrt(2x + 1) + sqrt(3x + 4) = sqrt(7x - 3)#

Good luck!