MHB Radical Equation Without Constant

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The discussion focuses on solving the equation involving three radicals: sqrt{2t + 5} - sqrt{8t + 25} + sqrt{2t + 8} = 0. Participants suggest rewriting the equation to isolate the radicals on one side, leading to the form sqrt{2t+5} + sqrt{2t+8} = sqrt{8t+25}. A key point is the importance of squaring both sides correctly, noting that squaring a sum involves a cross term, which must be accounted for. The conversation emphasizes the need to eliminate the radicals iteratively, as each squaring may introduce additional complexity. Overall, the problem requires careful manipulation of radicals to find all real number t values.
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Solve for all real number t values.

sqrt {2t + 5} - sqrt {8t + 25} + sqrt {2t + 8} = 0

I see there are no constants in this problem. I typically isolate the radical on one side of the equation and the constant (s) on the other side but there are 3 radicals on the left side. This is strange.

Can someone get me started?
 
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Rewrite as $\sqrt{2t+5}+\sqrt{2t+8}=\sqrt{8t+25}$. What do you get when you square both sides?
 
greg1313 said:
Rewrite as $\sqrt{2t+5}+\sqrt{2t+8}=\sqrt{8t+25}$. What do you get when you square both sides?

I had no idea that it is legal to move one radical over to the other side. When I square both sides, the radicals go away.
 
RTCNTC said:
I had no idea that it is legal to move one radical over to the other side. When I square both sides, the radicals go away.
At a guess you are forgetting about the cross term. [math](a + b)^2 \neq a^2 + b^2[/math]. It is [math](a + b)^2 = a^2 + 2ab + b^2[/math]. You are going to end up having to apply getting rid of the radicals two times. For each one you pick a term and put it on the RHS. Then square it.

-Dan
 
Very good. I will work on this later.
 
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