+129849 √(x + 7) - √(6 - x) = 1 add √(6 - x) to both sides to separate the radicals
√(x + 7) = 1 + √(6 - x) square both sides
x + 7 = 1 + 2 √(6 - x) + (6 -x) rearrange to get theradical on one side
2x =2 √(6 - x) divide both sides by 2
x = √(6 - x) square both sides, again
x^2 = 6 - x rearrange
x^2 + x - 6 = 0 factor
(x + 3) ( x - 2) = 0
And setting each factor to 0, we have that x = -3 or x = 2 ......we must check these in the original problem, as these types of equations often lead to "extraneous" solutions [due to the squaring of both sides]
Check -3 ... [√(-3 + 7) - √(6 - (-3)) = 1 ] = [2 - 3 = 1 ] is untrue
Check 2 ... [ √(2 + 7) - √(6 - 2) = 1 ] = [ 3 - 2 ] = 1 which is true
So x = 2 is the only solution.......here's a graph of both sides of the original problem.....
https://www.desmos.com/calculator/rmqhq2arba
Notice that the only solution occurs at (2, 1)
+129849 √(x + 7) - √(6 - x) = 1 add √(6 - x) to both sides to separate the radicals
√(x + 7) = 1 + √(6 - x) square both sides
x + 7 = 1 + 2 √(6 - x) + (6 -x) rearrange to get theradical on one side
2x =2 √(6 - x) divide both sides by 2
x = √(6 - x) square both sides, again
x^2 = 6 - x rearrange
x^2 + x - 6 = 0 factor
(x + 3) ( x - 2) = 0
And setting each factor to 0, we have that x = -3 or x = 2 ......we must check these in the original problem, as these types of equations often lead to "extraneous" solutions [due to the squaring of both sides]
Check -3 ... [√(-3 + 7) - √(6 - (-3)) = 1 ] = [2 - 3 = 1 ] is untrue
Check 2 ... [ √(2 + 7) - √(6 - 2) = 1 ] = [ 3 - 2 ] = 1 which is true
So x = 2 is the only solution.......here's a graph of both sides of the original problem.....
https://www.desmos.com/calculator/rmqhq2arba
Notice that the only solution occurs at (2, 1)
