View Full Version : Weird problem showed up on my math final today... help

06-03-2004, 11:58 AM
so yah, i know that the final is over, but this question is really bugging me. It was an equation and i can normally do these. So the equation was:

square root of (x^2 + 5) = x - 5

Now, what i did was square both sides, and then i added like terms and such to get down to:

10x = 20, so naturally i thought x was 2, and that was one of the choices (multiple choice test) so i put down that it was 2. I had some time left at the end of the test, so i was checking over, and i plugged 2 back into the equation, and i got:

3 = -3, not 3 = 3 like i should have if it was the solution. So then what i did was, i put:

y = square root of (x^2 + 5) and
y = x - 5

into my calculator as two separate equations and graphed them to see where they intersected, (their intersection is a solution) and when i used the function on the calculator that finds the intersection, i got an error. So, another one of the solutions was no solution, so i changed it to that. I was just wondering if anyone knew why the answer i got algebraiclly doesn't work? i mean all my algebra and arithmetic are correct, so it doesn't make much sense.

Any help would be great!!


Peter Tsai
06-03-2004, 10:05 PM

you will get two imaginary values for x ...

after moving all the terms to one side:

x^2 - x +10 = 0

then use the quadratic formula to solve for x

Peter Tsai
06-03-2004, 10:49 PM
sorry...i didnt see the squareroot in front of your problem


whenever you square both sides of the equation ... you lose the distinction of the possibility of one side being negative

for example x^2 = (-x)^2

but realistically "x" does not equal -x unless x = 0.

hope this helps

Peter Tsai
06-03-2004, 11:00 PM
if the problem was multiple choice ... then just plug in all the values for x given and if none of them work then the answer is no solution

06-04-2004, 04:11 AM
yah, i was thinking about that, but then i was thinking, that since i ended up with 3 = -3, that the 3, could actually be a -3, because the square root of 9 is a 3 and a -3, is this correct? i think i screwed myself over.


06-06-2004, 07:12 AM
here is a website explaining how to solve:

after squaring both sides and solving for x, you get x=2, which if you plug back into the equation, does not work, therefore the answer is "no solution"

06-06-2004, 10:26 AM
More often then not the answer is "C".

06-06-2004, 02:32 PM
The Question is:
square root of (x^2 + 5) = x - 5

You first Square both sides, then it comes out to be:
X^2+5 = X^2-10X+25
By Looking at both equations, you know that they're Parabolas.
Then You cancel out the X^2 by subtracting on both sides. And isolate the X.
Then it comes out to be:
10X = 20, which means X=2.
Now, take out your Graphic Calculator to see it on a graph.
Make sure it's in function mode. I have a TI-83+ calculator.
When you type it in the calulator, for the first Y= put in X^2+5, then for the next Y= put in X^2-10X+25.
Now go to Zoom Standard. Then Press Trace and plugg in 2 for X.
And click Zoom IN.
Now you can see where they intersect, X=2, Y=9.

06-06-2004, 02:39 PM
Now, take all the answers from the multiple choice and plugg them in, if nothing works. Then there is no solution.

06-07-2004, 03:59 AM
Ace has aced this test with his response above - no solution is the correct answer.

06-07-2004, 10:57 AM
Unfortunately, I'm much better at math than tennis.

Max G.
06-07-2004, 03:20 PM
Heh. Fortunately, I'm a much at math than tennis.

06-07-2004, 11:12 PM
I'm not sure about "no solution" being the right answer. I think the answer is no solution for the positive roots and x=2 for the negative roots. There are two curves when you graph an even root function, and the line y=x-5 intersects with the negative one.

Max G.
06-08-2004, 09:24 AM
True. It depends on whether the square root sign is interpreted as positive or negative, or just positive.

06-08-2004, 11:28 AM
no solution is the correct answer

James Brown
06-09-2004, 05:29 PM
what math are you taking? 436 or something?