Solve the equation x3 x2 5 Be sure to check for extraneous

Solve the equation (x-3) + (x+2) = 5. Be sure to check for extraneous solutions.

Solution

Given that

(x-3) + (x+2) = 5

  (x-3) = 5 - (x+2)

Squaring on both sides of equation

[ (x-3) ]² = [ 5 - (x+2) ]²

x - 3 = 5² + [ (x+2) ]² - 2 .5 .(x+2)

x - 3 = 25 + (x +2) - 10 (x+2)

  10 (x+2) = 25 +x +2-x+3

10 (x+2) = 30

  (x+2) = 30 / 10

  (x+2) = 3

x + 2 = 3²

x = 3² - 2

x = 7

Check for x = 7 in given equation

(x-3) + (x+2) = 5

  (7-3) + (7+2) = 5

4 + 9 = 5

2 + 3 =5

5 = 5

The solution is valid

A solution of a simplified version of an equation that does not satisfy the original equation,that solution is called extraneous solution.

Therefore,

x = 7 is not extraneous solution