algebra help

**Mat724** · May 29th 2012, 05:33 AM

an equation is shown.

[the square root of (x²-t²)] = 2t-x

If x and t are positive numbers that satisfy the equation above, what is the value of (x/t)?

**Reckoner** · May 29th 2012, 06:43 AM

Originally Posted by Mat724

[the square root of (x²-t²)] = 2t-x

If x and t are positive numbers that satisfy the equation above, what is the value of (x/t)?

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

$\Rightarrow x^2-t^2 = 4t^2-4tx+x^2$

$\Rightarrow 5t^2-4tx = 0$

$\Rightarrow t(5t - 4x) = 0$

$t > 0$ , so

$5t-4x = 0\quad\Rightarrow\quad4x=5t\quad\Rightarrow\quad \frac xt=\frac54$

**DeMath** · May 29th 2012, 06:49 AM

Hint

$\begin{gathered} \sqrt {{x^2} - {t^2}} = 2t - x \hfill \\ t\sqrt {{{\left( {\frac{x}{t}} \right)}^2} - 1} = 2t - x \hfill \\ \sqrt {{{\left( {\frac{x}{t}} \right)}^2} - 1} = 2 - \frac{x}{t} \hfill \\ \text{Let}\,\frac{x}{t} = a,\quad\text{domain}\colon\,|a| \geqslant 1 \hfill \\ \sqrt {{a^2} - 1} = 2 - a \hfill \\ {a^2} - 1 = {(2 - a)^2} \hfill \\{a^2} - 1 = 4 - 4a + {a^2} \hfill \\ 4a = 5 \hfill \\ a = \frac{5}{4} \hfill \\ \end{gathered}$

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