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
A:
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:
[tex]t=t^2[/tex]
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.