Finding x/t when √(x² − t²) = 2t − x

[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.

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

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.

$$x^2 - t^2 = \left(2t-x\right)^2 = 4t^2 - 4xt + x^2$$

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

$$4t^2 - x^2$$

which isn't correct, since

$$(2t-x)^2 \ne 4t^2 -x^2$$

Continuing on with the work,

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

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.