Help With Finding Real Roots

[FyreSakura]

HI! I need help answering this question for my Accel Algebra III homework:

Calculate the [real] root(s) or r(x)= sqrt(x-5) + 1

We did not learn how to do real roots and so I am not quite sure how to solve this. Also, I have tried looking online for help but I can't find anything.

Thanks for helping!
FyreSakura

 

[DrSteve]

The roots are the x-values where the function is 0:

\(\displaystyle \sqrt{x-5} + 1=0\)

Isolate the square root:

\(\displaystyle \sqrt{x-5}=-1\)

Square both sides:

\(\displaystyle x-5=1\)

Isolate x:

\(\displaystyle x=6\)

Make sure to check this answer back in the original equation.

\(\displaystyle \sqrt{6-5} + 1=0\)

\(\displaystyle \sqrt{1} + 1=0\)

\(\displaystyle 1+1=0\)

\(\displaystyle 2=0\)

This last equation is false which means the root we got was extraneous. Thus, we reject it. So there are no real roots.

 

[lookagain]

Calculate the [real] root(s) or r(x)= sqrt(x-5) + 1

FyreSakura

The roots are the x-values where the function is 0:

\(\displaystyle \sqrt{x-5} + 1=0\)

Isolate the square root:

\(\displaystyle \sqrt{x-5}=-1\)

FyreSakura,

you could stop at this point and declare that there are no real roots,
because a square root of a number cannot equal a negative number.

 

[FyreSakura]

Thank you so much! I understand now!

FyreSakura