SOLUTION: I am supposed to solve this: 3+ root symbol (z-8) = root symbol(z+7) I am not sure what to do here, any help would be appreciated! With explanation if possible!

Algebra ->  Radicals -> SOLUTION: I am supposed to solve this: 3+ root symbol (z-8) = root symbol(z+7) I am not sure what to do here, any help would be appreciated! With explanation if possible!      Log On


   

Question 465887: I am supposed to solve this:
3+ root symbol (z-8) = root symbol(z+7)
I am not sure what to do here, any help would be appreciated! With explanation if possible!
Found 2 solutions by solver91311, bucky:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Use sqrt(x) to indicate a radical. Everything inside the parentheses goes inside the radical.



Square both sides, using FOIL in the LHS.



Move all but the remaining radical to the RHS and collect like terms:



Multiply both sides by



Square both sides again:





Substitute this value back into the original equation and do the arithmetic to check the work.

John

My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
3%2Bsqrt%28z-8%29=+sqrt%28z%2B7%29
.
The next step is to square both sides. Squaring the left side is done by multiplying:
.
%283%2Bsqrt%28z-8%29%29%2A%283%2Bsqrt%28z-8%29%29
.
This can be done by using the FOIL method ... multiply First Terms, then Outside Terms, then Inside Terms, and finally Last Terms. When you do that you get:
.
%283%2A3%29%2B3%2Asqrt%28z-8%29+%2B3%2Asqrt%28z-8%29%2Bsqrt%28z-8%29%2Asqrt%28z-8%29
.
This simplifies to:
.
9+%2B+6%2Asqrt%28z-8%29%2B%28z-8%29 which further simplifies to:
.
9+%2B+6%2Asqrt%28z-8%29%2Bz+-8
.
and the +9 and -8 combines to +1 so the expression reduces to:
.
6%2Asqrt%28z-8%29+%2Bz+%2B+1
.
Now return to square the right side of the equation as follows:
.
sqrt%28z%2B7%29%2Asqrt%28z%2B7%29+=+z%2B7
.
So after squaring both sides the equation becomes:
.
6%2Asqrt%28z-8%29+%2B+z+%2B+1+=+z+%2B+7
.
Subtract z from both sides reduces this to:
.
6%2Asqrt%28z-8%29+%2B+1+=+7
.
Next subtract 1 from both sides to get:
.
6%2Asqrt%28z-8%29+=+6
.
Divide both sides by 6:
.
sqrt%28z-8%29+=+1
.
Square both sides:
.
z-8+=+1
.
Finally add 8 to both sides and the result is:
.
z+=+9
.
That's the answer. You can check this by returning to the original equation, substituting 9 for z, and working it out to see that both sides are equal. In other words, start with:
.
3%2Bsqrt%28z-8%29=+sqrt%28z%2B7%29
.
Substitute 9 for z:
.
3+%2B+sqrt%289-8%29=sqrt%289%2B7%29
.
Combine the numbers under each radical to get:
.
3+%2B+sqrt%281%29=sqrt%2816%29
.
This reduces to:
.
3+%2B+1+=+4
.
And this is true (4 does equal 4) so our answer of z = 9 does satisfy the original equation.
.
Hope this helps you to understand that squaring radicals is a way to solve problems such as these.