Hi, first post:
I'm helping a student work a radical equation and have a question. Here is the problem:
√2X-5 = 1 + √X - 3 (2X - 5 AND X - 3 ARE ALL UNDER THE RADICAL SIGN)
Step 2: (√2x – 5)2 = (1 + √x-3)2
Step 3: 2x – 5 = 1 = 2√x-3 + (x -3 ) à My question is HOW did they get the 2 in front of the radical sign!!!
Step 4: x – 3 = 2√x-3 à so I’m really lost on this one.
The answer is x = 7 or x = 3 but WHAT did they do between step 2 and step 3 to get the 2 in front of the radical sign?
Thanks.
I'll do this question step by step so you have an example how to do the next 166 questions of the same type:
1. In general: To get rid of the square-root you have to square - but correctly:
(a * b)² = a² * b²
(a+b)² = a² + 2ab + b²
(a-b)² = a² - 2ab + b²
2. The tactics in solving such an equation is: Square - isolate remainings roots - square - isolate .... until there isn't any root left:
3. Your question:
........ square both sides
........ collect like terms and isolate the root on the RHS of the equation
........ square both sides
........ collect like terms
........ use the formula to solve this quadratic equation. I've got:
Since squaring is not an equivalent transformation you have to plug in the results into the original equation and check if the solutions satisfy the equation:
1.
2.
Obviously both results are solutions of the equation.