Solve SAT Math Question: \frac{x}{t} with x and t Positive

[JBD2]

Homework Statement

\sqrt{x^{2}-t^{2}}=2t-x

If x and t are positive numbers that satisfy the equation above, what is the value of \frac{x}{t}?

2. The attempt at a solution
x^{2}-t^{2}=4t^{2}-x^{2}

x^{2}+x^{2}=4t^{2}+t^{2}

2x^{2}=5t^{2}

I'm unsure of what to do next.

 

[gabbagabbahey]

Divide both sides of the equation by 2t^2 and then take the square root.

 

[JBD2]

Do you mean divide:

2x^{2}=5t^{2} by 2t^{2}? How does that work?

 

[gabbagabbahey]

Yes, dividing both sides of the equation gives you:

\frac{2x^2}{2t^2}=\frac{5t^2}{2t^2} \Rightarrow \frac{x^2}{t^2}=\frac{5}{2}

Then take the square root of both sides of the equation, that should give you x/t=...

 

[JBD2]

The answer is apparently \frac{5}{4} though, and I don't understand where the 2t² comes from.

 

[statdad]

Divide by 2 t^2 to create an equation in which one side is constant, the other involving only x \text{ and } t.

 

[JBD2]

How do I get 5/4 then?

 

[statdad]

I'm not sure, because looking back at your earlier posts (which I didn't do the first time) I think you have an early error.
Here is your original equation.

<br /> \sqrt{x^{2}-t^{2}}=2t-x<br />

Notice that we cannot have x = 0, since \sqrt{-t^2} is not a real number. Since you want the value of x/t we don't need to consider t = 0 either.

The first step in the solution is to square each side.

<br /> x^2 - t^2 = \left(2t-x\right)^2 = 4t^2 - 4xt + x^2<br />

The right hand side on your first step of the solution is only

<br /> 4t^2 - x^2<br />

which isn't correct, since

<br /> (2t-x)^2 \ne 4t^2 -x^2<br />

Continuing on with the work,

<br /> \begin{align*}<br /> x^2 - t^2 &amp; = 4t^2 - 4xt + x^2 \\<br /> -t^2 &amp; = 4t^2 - 4xt \\<br /> -5t^2 &amp; = -4xt <br /> \end{align*}<br />

Remembering you want to reach a point in which you have x/t on one side,
and a constant on the other, what do you need to do to the final equation here to reach it? (It is at this point that knowing t cannot equal zero is important.)

 

[JBD2]

-5 = \frac{-4x}{t}
\frac{5}{4} = \frac{x}{t}

Oh ok I get it now, thanks so much.