Find all real solutions

• May 11th 2009, 08:35 PM
aperkin2
Find all real solutions
Hi! I am studying for a final and have gotten stuck (over & over again!)
On this certain problem, I am supposed to find all real solutions for the square root of x + 3 + the square root of x + 8= 1
So far, I have done...
square root of x + 3 = 1 - the square root of x + 8
(square root of x + 3 )^2= (1 - the square root of x + 8)^2

And, I'm not really sure where to go from here. Any help would be greatly appreciated!

Thanks,
Ashley
• May 11th 2009, 08:54 PM
Chris L T521
Quote:

Originally Posted by aperkin2
Hi! I am studying for a final and have gotten stuck (over & over again!)
On this certain problem, I am supposed to find all real solutions for the square root of x + 3 + the square root of x + 8= 1
So far, I have done...
square root of x + 3 = 1 - the square root of x + 8
(square root of x + 3 )^2= (1 - the square root of x + 8)^2

And, I'm not really sure where to go from here. Any help would be greatly appreciated!

Thanks,
Ashley

From there, expand and simplify:

$\left(\sqrt{x+3}\right)^2=\left(1-\sqrt{x+8}\right)^2\implies x+3=1-2\sqrt{x+8}+x+8$ $\implies 2\sqrt{x+8}=6\implies \sqrt{x+8}=3\implies x=1$

But when you plug in $x=1$ into the expression, you end up with $\sqrt{4}+\sqrt{9}=1\implies 5=1$ which is impossible. Therefore, there is no real solution.

Does this make sense?
• May 11th 2009, 09:12 PM
aperkin2
Hi! I do understand your answer. I didn't know that (1 - the square root of x + 8)^2 would give me 1-2 the square root of x + 8) + x+ 8. I am sure a lot of my problems stem from having trouble with the basics. Thank you so much for your help!

Ashley