# Find a radical equation that has solution x=5 and extraneous solutions x=-2 and x=-1

• Sep 6th 2011, 04:10 PM
anna123
Find a radical equation that has solution x=5 and extraneous solutions x=-2 and 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
• Sep 6th 2011, 04:32 PM
pickslides
Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and
How did you find it?

Could you start by expanding (x-5)(x+2)(x+1) ?
• Sep 6th 2011, 07:03 PM
anna123
Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and
Yes, I started by expanding (x-5)(x+2)(x+1)
• Sep 6th 2011, 07:04 PM
pickslides
Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and
Then what did you do?
• Sep 6th 2011, 07:10 PM
anna123
Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and
Look at the attached document
[IMG]file:///C:/Users/MEYYAP%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
Attachment 22200
• Sep 6th 2011, 08:01 PM
SammyS
Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and
Quote:

Originally Posted by anna123
Yes, I started by expanding (x-5)(x+2)(x+1)

Expand that & set it equal to zero, then subtract a perfect square which is positive at x = -1 & x = -2, but is zero at x = 5. The simplest one of those is (x-5)^2.

Then take the square root of both sides of the resulting equation.
• Sep 7th 2011, 03:08 PM
anna123
Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and
After expanding, I am getting x^3-2x^2-13x-10=0. Are you saying I have to subtract (x-5)^2 from both sides?

x^3-2x^2-13x-10-(x-5)^2 = 0 - (x-5)^2?????
• Sep 7th 2011, 03:15 PM
pickslides
Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and
Yep, now expand $-(x-5)^2$ on the LHS and group like terms, then take the square root of both sides, then you are finished.
• Sep 7th 2011, 04:53 PM
anna123
Re: Find a radical equation that has solution x=5 and extraneous solutions x=-2 and
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?
• Sep 7th 2011, 07:16 PM
SammyS
Re: Find a radical equation that has solution x=5 and extraneous solutions x=-2 and
Quote:

Originally Posted by anna123
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 $(\sqrt{x+2}-5)(\sqrt{x+2}+3)=0\,,$ 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, $(\sqrt{x+2}+3)$ 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:

$x-2\sqrt{x+2}=13$ or any equivalent equation.

For a solution x = 5, and an extraneous solution of x = -2, try: $(\sqrt{x+11}-4)(\sqrt{x+11}+3)=0\,.$

However, I find it very difficult to get one valid solution AND two extraneous solutions.
• Sep 8th 2011, 03:22 AM
Opalg
Re: Find a radical equation that has solution x=5 and extraneous solutions x=-2 and
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 $x^3-2x^2 - 13x - 10=0.$ 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 $x^2 = \tfrac12(x^3-13x-10).$ Now take the square root to get $\boxed{x = \sqrt{\tfrac12(x^3-13x-10)}}.$