SOLUTION: how would you solve the following equations? Show different ways to solve each. 1. {{{root(3,x-5)}}}{{{""=""}}}{{{sqrt(3)}}} 2. {{{3/sqrt(x-3)}}}{{{""=""}}}{{{4/sqrt(2x+3)}}} 3.

Algebra ->  Square-cubic-other-roots -> SOLUTION: how would you solve the following equations? Show different ways to solve each. 1. {{{root(3,x-5)}}}{{{""=""}}}{{{sqrt(3)}}} 2. {{{3/sqrt(x-3)}}}{{{""=""}}}{{{4/sqrt(2x+3)}}} 3.      Log On


   

Question 697655: how would you solve the following equations? Show different ways to solve each.
1. root%283%2Cx-5%29%22%22=%22%22sqrt%283%29
2. 3%2Fsqrt%28x-3%29%22%22=%22%224%2Fsqrt%282x%2B3%29
3. root%284%2C2x-5%29%22%22=%22%22sqrt%28x%2B1%29
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
1. root%283%2Cx-5%29%22%22=%22%22sqrt%283%29

Cube both sides:

%28root%283%2Cx-5%29%29%5E3%22%22=%22%22%28sqrt%283%29%29%5E3

x-5%22%22=%22%22%28sqrt%283%29%29%5E%282%2B1%29

x-5%22%22=%22%22%28sqrt%283%29%29%5E2%2Asqrt%283%29
x-5%22%22=%22%223%2Asqrt%283%29

   x%22%22=%22%225%22%22%2B%22%223sqrt%283%29

There is no need to check that because we raised variables
only to an ODD power.  It is only when you raise variables
to an even power that you might get extraneous solutions.



2.  3%2Fsqrt%28x-3%29%22%22=%22%224%2Fsqrt%282x%2B3%29

Cross-multiply

3%2Asqrt%282x%2B3%29%22%22=%22%224%2Asqrt%28x-3%29

Square both sides:

%283%2Asqrt%282x%2B3%29%29%5E2%22%22=%22%22%284%2Asqrt%28x-3%29%29%5E2

3%5E2%28sqrt%282x%2B3%29%29%5E2%22%22=%22%224%5E2%28sqrt%28x-3%29%29%5E2

9%282x%2B3%29%22%22=%22%2216%28x-3%29

18x%2B27%22%22=%22%2216x-48

2x%22%22=%22%22-75

   x%22%22=%22%22-75%2F2

We must check this because we raised variables to an
EVEN power, and this may be an extraneous solution.

3%2Fsqrt%28x-3%29%22%22=%22%224%2Fsqrt%282x%2B3%29
3%2Fsqrt%28-75%2F2-3%29%22%22=%22%224%2Fsqrt%282-75%2F2%29%2B3%29

We need go no further because there can be no real
solution if there is a negative number under an EVEN
root, such as a square root, 4th root, 6th root, etc.

This equation has no real solution.


3. root%284%2C2x-5%29%22%22=%22%22sqrt%28x%2B1%29

Raise both sides to the 4th power:

%28root%284%2C2x-5%29%29%5E4%22%22=%22%22%28sqrt%28x%2B1%29%29%5E4

2x-5%22%22=%22%22%28sqrt%28x%2B1%29%29%5E2%2A%28sqrt%28x%2B1%29%29%5E2

2x-5%22%22=%22%22%28x%2B1%29%28x%2B1%29

2x-5%22%22=%22%22x%5E2%2B2x%2B1

-6%22%22=%22%22x%5E2

This can have no real solution.

Edwin