How do you solve #x+5=sqrt(17+x)?#?

2 Answers
Jun 22, 2018

Answer:

#x=-1#

Explanation:

#x+5=sqrt(17+x)#

Square both sides
#x^2+10x+25=17+x#

Subtract to make one side 0
#x^2+9x+8=0#

Factor
#(x+8)(x+1)=0#

Solve for x:
#x+8=0 or x+1=0#
#x=-8 or x=-1#

Since we squared both sides, we must check for extraneous solutions:
#-1+5=sqrt(17-1)#
#4=4#, #x=-1#

#-8+5=sqrt(17-8)#
#-3=sqrt(9)#
By convention, the #sqrt()# symbol always refers to the positive square root, so:
#-3!=3#, #x!=-8#

Therefore, #x=-1#

Jun 22, 2018

Answer:

#x = -1#

Explanation:

#x + 5 = sqrt(17+x)#

First, square both sides:
#(x+5)^color(blue)2 = (sqrt(17+x))^color(blue)2#

#(x+5)^2 = 17+x#

We know that any binomial in the form of:
#(x+y)^2 = x^2 + 2xy + y^2#, so the left hand side becomes:
#x^2 + 10x + 25#

Put that back into the equation:
#x^2 + 10x + 25 = 17 + x#

Subtract #color(blue)(17)# and #color(blue)x# from both sides:
#x^2 + 9x + 8 = 0#

This is in standard form, or #color(red)(a)x^2 + color(green)(b)x + color(magenta)(c)#. In this form, we can factor it with two numbers that:

  1. Multiply up to #color(red)(a)* color(magenta)(c) = color(red)(1) * color(magenta)(8) = 8#
  2. Add up to #color(green)(b)#, or #color(green)9#

Those two numbers are #color(blue)8# and #color(blue)1#, as:

  1. #1 * 8 = 8#
  2. #1 + 8 = 9#

Therefore, the factored form is:
#(x + 8)(x + 1) = 0#

Since when the expressions multiply to equal zero, that means each of the expressions equal zero. So we can set up two equations:
#x+8 = 0# and #x+1 = 0#

Subtract #color(blue)8# in the first equation, subtract #color(blue)1# in the second equation:
#x = -8# and #x = -1#

#--------------------#

Now we have to check if both solutions are really solutions by plugging them into the original formula. Let's check #color(blue)(-8)# first:

#x + 5 = sqrt(17+x)#

#-8 + 5 = sqrt(17-8)#

#-3 = sqrt9#

#-3 != 3#

No, this is not true, so #color(blue)(-8)# is NOT a solution! Now check #color(blue)(-1)#:
#-1 + 5 = sqrt(17-1)#

#4 = sqrt16#

#4 = 4#

Yes, this is true, so #color(blue)(-1)# is the ONLY solution!

Hope this helps!