How do I get $x$ by itself in this equation? This is work related not school related. Thanks!
$x^2 +(tx)^2 = 1.333^2$
How do I get $x$ by itself in this equation? This is work related not school related. Thanks!
$x^2 +(tx)^2 = 1.333^2$
Note that $x^2 +(tx)^2 = x^2 +t^2 x^2 = (1+t^2)x^2$.
Hence, $x^2 +(tx)^2 = 1.333^2$ is the same as $(1+t^2)x^2 = 1.333^2$.
If $t^2 \neq -1$, we can dividing throughout by $1+t^2$, to get $x^2 = \dfrac{1.333^2}{1+t^2}$
Taking the square-root on both sides, we get $x = \pm \sqrt{\dfrac{1.333^2}{1+t^2}} = \pm \dfrac{\sqrt{1.333^2}}{\sqrt{1+t^2}} = \pm \dfrac{1.333}{\sqrt{1+t^2}}$
$x^2 + t^2 x^2 = 1.333^2$
$x^2(1 + t^2) = 1.333^2$
$x^2 = \frac{1.333^2}{1 + t^2}$
$x = \sqrt{\frac{1.333^2}{1 + t^2}}$
Remember the negative square root.
As noted by Hurkyl, you need $1 + t^2 \neq 0$. However, note that if $1 + t^2 = 0$, then $t^2 = -1$. Hence $x^2 + t^2x^2 = x^2 - x^2 \neq 1.333^2$. If $t \in \mathbb{R}$, then $1 + t^2 \neq 0$ .
QED