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How do I get $x$ by itself in this equation? This is work related not school related. Thanks!

$x^2 +(tx)^2 = 1.333^2$

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    Maybe $1.333$ was meant for $1.\hat 3$?2012-05-22

3 Answers 3

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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 $x = \pm \sqrt{\dfrac{1.333^2}{1+t^2}} = \pm \dfrac{\sqrt{1.333^2}}{\sqrt{1+t^2}} = \pm \dfrac{1.333}{\sqrt{1+t^2}}$

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    @TMM Thanks. Have updated it.2012-05-22
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$x^2 + t^2 x^2 = 1.333^2$

$x^2(1 + t^2) = 1.333^2$

$x^2 = \frac{1.333^2}{1 + t^2}$

$x = \sqrt{\frac{1.333^2}{1 + t^2}}$

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 $1 + t^2 \neq 0$ .

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    ... assuming $1 + t^2$ is nonzero. This is fine, of course, if we're assuming $t$ is a real number.2012-05-22
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  1. We can turn $(tx)^2$ into $t^2x^2$ from the Exponent Laws, so we now have $x^2+x^2t^2=1.333^2$
  2. I then like to factor out like terms, so we get $x^2(1+t^2)=1.333^2$
  3. We then divide both sides by $1+t^2$ so we have only have $x$ on the left hand side. To do this, we must make sure $1+t^2 \ne0$ to avoid division by zero. Assuming $t$ is a real number, this is not an issue because for $1+t^2$ to equal $0$, $t$ must be $\pm i$ where $i$ is the imaginary unit (which is not a real number). We now have $x^2=\frac{1.333^2}{1+t^2}$
  4. Taking the positive and negative square roots of the sides, we find $x=\pm \sqrt{\frac{1.333^2}{1+t^2}}$
  5. Again using exponent laws, we may simplify the RHS: $x=\pm \frac{\sqrt{1.333^2}}{\sqrt{1+t^2}}=\pm\frac{1.333}{\sqrt{1+t^2}}$

QED