J Jaskaran Junior Member Joined May 5, 2006 Messages 67 Mar 20, 2007 #1 I have a problem, √(7-x) - 5 = √(6-x) I'm confused on how to approach it, should I add the five to the other side... But that would give me (after squaring the equation) 7 - x = 6 - x +10√(6-x) +25 Where to from there?!
I have a problem, √(7-x) - 5 = √(6-x) I'm confused on how to approach it, should I add the five to the other side... But that would give me (after squaring the equation) 7 - x = 6 - x +10√(6-x) +25 Where to from there?!
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,550 Mar 20, 2007 #2 Isolate the radical, and square everything again. Eliz.
J Jaskaran Junior Member Joined May 5, 2006 Messages 67 Mar 20, 2007 #3 So it's: 7 - x = 6 - x +10√(6-x) +25 -10√(6-x) = 24 100(6-x) = 240 6 - x = 24 x = -18 ?
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,550 Mar 21, 2007 #4 Jaskaran said: -10√(6-x) = 24 100(6-x) = 240 Click to expand... How did you get that 24<sup>2</sup> equalled 240...? Eliz.
Jaskaran said: -10√(6-x) = 24 100(6-x) = 240 Click to expand... How did you get that 24<sup>2</sup> equalled 240...? Eliz.
