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

In summary, the given equation is \sqrt{x^{2}-t^{2}}=2t-x and we are asked to find the value of \frac{x}{t} if x and t are positive numbers that satisfy the equation. After squaring both sides and simplifying, we reach an equation of -5t^2 = -4xt. Remembering that t cannot equal 0, we can divide both sides by -4t to get the value of \frac{x}{t} which is \frac{5}{4}.
  • #1
JBD2
61
0

Homework Statement


[tex]\sqrt{x^{2}-t^{2}}=2t-x[/tex]

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

2. The attempt at a solution
[tex]x^{2}-t^{2}=4t^{2}-x^{2}[/tex]

[tex]x^{2}+x^{2}=4t^{2}+t^{2}[/tex]

[tex]2x^{2}=5t^{2}[/tex]

I'm unsure of what to do next.
 
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  • #2
Divide both sides of the equation by [itex]2t^2[/itex] and then take the square root.
 
  • #3
Do you mean divide:

[tex]2x^{2}=5t^{2}[/tex] by [tex]2t^{2}[/tex]? How does that work?
 
  • #4
Yes, dividing both sides of the equation gives you:

[tex]\frac{2x^2}{2t^2}=\frac{5t^2}{2t^2} \Rightarrow \frac{x^2}{t^2}=\frac{5}{2}[/tex]

Then take the square root of both sides of the equation, that should give you x/t=...
 
  • #5
The answer is apparently [tex]\frac{5}{4}[/tex] though, and I don't understand where the 2t² comes from.
 
  • #6
Divide by [tex] 2 t^2 [/tex] to create an equation in which one side is constant, the other involving only [tex] x \text{ and } t [/tex].
 
  • #7
How do I get 5/4 then?
 
  • #8
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.

[tex]
\sqrt{x^{2}-t^{2}}=2t-x
[/tex]

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

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

[tex]
x^2 - t^2 = \left(2t-x\right)^2 = 4t^2 - 4xt + x^2
[/tex]

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

[tex]
4t^2 - x^2
[/tex]

which isn't correct, since

[tex]
(2t-x)^2 \ne 4t^2 -x^2
[/tex]

Continuing on with the work,

[tex]
\begin{align*}
x^2 - t^2 & = 4t^2 - 4xt + x^2 \\
-t^2 & = 4t^2 - 4xt \\
-5t^2 & = -4xt
\end{align*}
[/tex]

Remembering you want to reach a point in which you have [tex] x/t [/tex] 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 [tex] t [/tex] cannot equal zero is important.)
 
  • #9
[tex]-5 = \frac{-4x}{t}[/tex]
[tex]\frac{5}{4} = \frac{x}{t}[/tex]

Oh ok I get it now, thanks so much.
 

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

1. What does the expression "x/t" mean in this equation?

The expression "x/t" in this equation represents the division of the variable x by the variable t. This is also known as a fraction or ratio.

2. How do I solve for the value of x in this equation?

To solve for x, you will need to isolate it on one side of the equation. To do this, multiply both sides by t. This will cancel out the t on the denominator of the fraction, leaving you with x on the numerator. The final step is to simplify the expression by dividing both sides by the coefficient of x, if there is one.

3. Can I use any value for x and t in this equation?

Yes, as long as x and t are both positive, you can use any value for these variables in the equation. However, keep in mind that the solution will be different depending on the values you choose.

4. What if x is equal to 0?

If x is equal to 0, then the entire expression becomes 0/t, which simplifies to 0. This means that the solution to the equation will also be 0, regardless of the value of t.

5. Can I use this equation to solve for t instead of x?

Yes, you can use this equation to solve for t by isolating t on one side of the equation. To do this, multiply both sides by t and then divide both sides by x. This will cancel out the x on the numerator of the fraction, leaving you with t on the denominator.

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