SOLUTION: Solve the equation sqrt (2x + 5) - sqrt (x - 2) = 3

Algebra ->  Expressions-with-variables -> SOLUTION: Solve the equation sqrt (2x + 5) - sqrt (x - 2) = 3      Log On


   

Question 172937This question is from textbook Introductory Algebra
: Solve the equation
sqrt (2x + 5) - sqrt (x - 2) = 3 This question is from textbook Introductory Algebra

Found 2 solutions by monika_p, Mathtut:
Answer by monika_p(71) About Me  (Show Source):
You can put this solution on YOUR website!
if you have variable inside a square root you have to do the opposit operation :squaring to "take out" x from root
[example %28sqrt%283%29%29%5E2=%283+%5E%281%2F2%29%29%5E2=3%5E%28%281%2F2%29%2A2%29=3%5E1=3]
sqrt%282x%2B5%29+-sqrt%28x-2%29=3 square both sides of the equation
%28sqrt%282x%2B5%29%29%5E2+-%28sqrt%28x-2%29%29%5E2=3%5E2
2x+5-(x-2)=9
x=2 answer

Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
although the answer is partially correct shown by monika ..she is missing a solution and got fortunate(or unfortunate depending on how you see it) to find the solution of 2. Why?? because when you do one thing to one side of the equation you have to do the same to the other....when you have one term to square on each side of the equals sign it is easy but when there are two terms which are added or subtracted......they must be taken as a whole and squared....you CANNOT square each term seperately unless the terms are multiplied together....with that in mind lets proceed to solve the problem. It is an easy mistake to make and is made quite often.
:
sqrt%282x%2B5%29-sqrt%28x-2%29=3given
:
%28sqrt%282x%2B5%29-sqrt%28x-2%29%29%5E2=3%5E2radicalize by squaring both sides
:
2x%2B5-2%28sqrt%28%282x%2B5%29%28x-2%29%29%29%2Bx-2=9multiplying out the squares
:
3x%2B3-2%28sqrt%28%282x%2B5%29%28x-2%29%29%29=9 combining like terms on left side of eq
:
-2%28sqrt%28%282x%2B5%29%28x-2%29%29%29=6-3xcombining like terms in the equation
:
%28-2%28sqrt%28%282x%2B5%29%28x-2%29%29%29%29%5E2=%286-3x%29%5E2squaring both sides to radicalize again
:
4%282x%2B5%29%28x-2%29=36-36x%2B9x%5E2squaring left removed radical...mult.right side
:
4%282x%5E2-4x%2B5x-10%29=36-36x%2B9x%5E2multiplied part of left side
:
8x%5E2-16x%2B20x-40=36-36x%2B9x%5E2multiplied out rest of left side
:
x%5E2-40x%2B76=0 combined like terms
:
answerssystem%28x=2%2Cx=38%29
:
:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-40x%2B76+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-40%29%5E2-4%2A1%2A76=1296.

Discriminant d=1296 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--40%2B-sqrt%28+1296+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-40%29%2Bsqrt%28+1296+%29%29%2F2%5C1+=+38
x%5B2%5D+=+%28-%28-40%29-sqrt%28+1296+%29%29%2F2%5C1+=+2

Quadratic expression 1x%5E2%2B-40x%2B76 can be factored:
1x%5E2%2B-40x%2B76+=+%28x-38%29%2A%28x-2%29
Again, the answer is: 38, 2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-40%2Ax%2B76+%29