MHB How to Solve an Equation with Square Roots?

AI Thread Summary
To solve the equation \(\sqrt{x}+\sqrt{x+8}=8\), first rearrange it to \(\sqrt{x+8}=8-\sqrt{x}\). Squaring both sides gives \(x+8=64-16\sqrt{x}+x\), leading to the equation \(16\sqrt{x}=56\). Dividing by 8 simplifies this to \(2\sqrt{x}=7\), resulting in \(\sqrt{x}=\frac{7}{2}\) and \(x=\frac{49}{4}\). Verification shows that this solution satisfies the original equation.
magda21
Messages
3
Reaction score
0
Please Help me solve it \[ \sqrt{x}+\sqrt{x+8}=8 \] thanks
 
Mathematics news on Phys.org
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. :)
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top