Find the non-extraneous solutions of the square root of the quantity x plus 6 minus 5 equals quantity x plus 1. x = −2 x = −6 and x = −5 x = −6 and x = −2 x = 2

Accepted Solution

The radical equation is [tex] \sqrt{x+6}-5=x+1 [/tex].

i) We first isolate the square root, adding 5 to both sides of the equation: 
                                        [tex] \sqrt{x+6}=x+6. [/tex]

ii) Here let's substitute x+6 with t. Doing so we have:
                                         [tex] \sqrt{t}=t. [/tex]

Squaring both sides, we get:

iii) Collecting the variables on the same side, and factorizing t we have:
                                               [tex]t(t-1)=0[/tex], which yields

                              t=0    or       t=1.

Now we solve for x in x+6=t:

x+6=0 ⇒x=-6       and x+6=1⇒x=-5.

iv) Now we check these values in the original equation  [tex] \sqrt{x+6}=x+6[/tex] :

a) [tex] \sqrt{-6+6}=-6+6. [/tex] ⇒ 0=0 ; Correct.

b) [tex] \sqrt{-5+6}=-5+6. [/tex] ⇒ 1=1 ; Correct.

Answer: x = −6 and x = −5