Radical Equation Without Constant

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SUMMARY

The discussion focuses on solving the radical equation sqrt{2t + 5} - sqrt{8t + 25} + sqrt{2t + 8} = 0 for real number t values. Participants emphasize the importance of isolating radicals and correctly applying the squaring technique to eliminate them. A key point is the necessity to account for cross terms when squaring both sides of an equation, as highlighted by the formula (a + b)^2 = a^2 + 2ab + b^2. The conversation concludes with a consensus on the need to apply the squaring method multiple times to fully resolve the equation.

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mathdad
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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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