Question

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

Answer 1

The radical equation is $\sqrt{x+6}-5=x+1$.


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

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

Squaring both sides, we get:
                       
                                               $t=t^2$

iii) Collecting the variables on the same side, and factorizing t we have:
 
                                               $t(t-1)=0$, 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  $\sqrt{x+6}=x+6$ :

a) $\sqrt{-6+6}=-6+6.$ ⇒ 0=0 ; Correct.


b) $\sqrt{-5+6}=-5+6.$ ⇒ 1=1 ; Correct.



Answer: x = −6 and x = −5 

Answer 2

Answer:

x = −6 and x = −7