How did you find it?
Could you start by expanding (x-5)(x+2)(x+1) ?
I found a radical equation, the equation is
(Square root of(2x^3-2x+16))= (2x+6).
When I check my equation to make sure the above equation generates
extraneous solution x=-1 and x=-2. I get all the three x=5, x=-1 and
x=-2 as solutions, all three satisfy my equation.
Can anybody comment on where I am going wrong
Look at the attached document
[IMG]file:///C:/Users/MEYYAP%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]Radical.doc
Thank you for your help!
I did subtract (x-5) from both sides and I ended up with the equation [IMG]file:///C:/Users/MEYYAP%7E1/AppData/Local/Temp/moz-screenshot.png[/IMG](x-5)= square root of (3x^2+3x+35-x^3).
Can you explain your reasoning behind why you wanted me to subtract x-5 from both sides?
Oh fiddlesticks !!! Now I see the problem with that!!! -(x-5)^2 isn't a perfect square.
Starting from scratch:
I like to take something like then multiply this out. The resulting equation will be solved by isolating the radical & squaring both sides. But if you look the above equation, the second factor, can not be zero, because the principal square root is not negative. But if you set this factor to zero, subtract 2 then square it, you get x+2 = 9 → x = 7, which will be an extraneous solution, if the following equation is solved conventionally. The other factor gives x = 23.
Multiplying the above equation out, gives:
or any equivalent equation.
For a solution x = 5, and an extraneous solution of x = -2, try:
However, I find it very difficult to get one valid solution AND two extraneous solutions.
I think this problem is easier than you are trying to make it. Expanding (x-5)(x+2)(x+1) as pickslides suggested, you get the equation You want to "radicalise" this in a way that works for the positive solution x=5 but not for the two negative solutions. So write it as Now take the square root to get