How to Solve an Equation with Square Roots?

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Discussion Overview

The discussion revolves around solving the equation \(\sqrt{x}+\sqrt{x+8}=8\). Participants explore various methods for manipulating the equation, addressing the squaring of both sides and the verification of solutions.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant requests help in solving the equation involving square roots.
  • Another participant suggests rearranging the equation to isolate one of the square roots before squaring both sides.
  • A subsequent reply attempts to solve the equation but questions their own steps after arriving at a value for \(x\).
  • Another participant corrects the squaring process and emphasizes the importance of checking for extraneous solutions.
  • A later reply provides a solution for \(x\) and expresses gratitude for the assistance.
  • One participant confirms the solution works in the original equation, indicating satisfaction with the process.

Areas of Agreement / Disagreement

Participants generally agree on the method of isolating square roots and squaring both sides, but there is a lack of consensus on the correctness of the initial steps taken by one participant.

Contextual Notes

There are unresolved aspects regarding the verification of solutions and the potential for extraneous solutions arising from squaring the equation.

Who May Find This Useful

Readers interested in solving equations involving square roots, particularly in a homework context, may find this discussion beneficial.

magda21
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Please Help me solve it \[ \sqrt{x}+\sqrt{x+8}=8 \] thanks
 
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Hello, and welcome to MHB! :)

I would begin by arranging as follows:

$$\sqrt{x+8}=8-\sqrt{x}$$

I really just want to arrange it so that there isn't two radicals on the same side. Now, what do you get when you square both sides?
 
x+8=64-x
2x=56
x=28
what I'm doing wrong?
 
Ah, you are not squaring the RHS correctly. Let's go back to:

$$\sqrt{x+8}=8-\sqrt{x}$$

Now, recall that:

$$(a+b)^2=a^2+2ab+b^2$$

And so, when we square both sides of our equation (bearing in mind that we must check for extraneous solutions), we get:

$$x+8=64-16\sqrt{x}+x$$

Collecting like terms, we can arrange this as:

$$16\sqrt{x}=56$$

Divide through by 8:

$$2\sqrt{x}=7$$

Can you proceed?
 
\[ \sqrt{x}=\frac{7}{2} \]
x=\frac{49}{4}
I see, thank you so much
 
Yes, and once we verifiy is works in the original equation, which it does, then we're done. :)
 

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