SOLUTION: Happy Easter! Equations with two radical terms. 1.) 3 + square root of z-6= square root of z+9 2.) If g(x)= square root of x + square root of x-5, find any x for which g(x

Algebra ->  Radicals -> SOLUTION: Happy Easter! Equations with two radical terms. 1.) 3 + square root of z-6= square root of z+9 2.) If g(x)= square root of x + square root of x-5, find any x for which g(x      Log On


   

Question 1026785: Happy Easter!
Equations with two radical terms.
1.) 3 + square root of z-6= square root of z+9
2.) If g(x)= square root of x + square root of x-5, find any x for which g(x)=5
3.) square root of a= -1
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
1.) 3%2B+sqrt%28z-6%29+=+sqrt%28z%2B9%29
==> 9%2B6sqrt%28z-6%29+%2B+z-6+=+z%2B9 after squaring both sides
==> 6sqrt%28z-6%29+=+6 after simplifying,
==> sqrt%28z-6%29+=+1
==> z-6 = 1 after squaring both sides.
==> z = 7.
Substituting z = 7 into the original equation to check if it is extraneous, it easily seen to satisfy the original equation, and so is the only solution.
2.) sqrt%28x%29+%2B+sqrt%28x-5%29+=+5
<==> sqrt%28x-5%29+=5-sqrt%28x%29
==> x-5+=+25+-+10sqrt%28x%29+%2B+x after squaring both sides,
==> 10sqrt%28x%29+=+30 after simplifying,
==> sqrt%28x%29+=+3 ==> x = 9.
Substituting x = 9 into the original equation to check if it is extraneous, it easily seen to satisfy the original equation, and so is the only solution.