2.1. Given $$ \frac{4 x-3}{x^{2}}=m $$, if -4 is one of the roots of the equation, determine the value of $$ m $$ and the other root.

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Solve the equation $$ 4x-3=m*x^{2} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$4x-3=mx^{2}$$
- step1: Move the expression to the left side:
  $$4x-3-mx^{2}=0$$
- step2: Rewrite in standard form:
  $$-mx^{2}+4x-3=0$$
- step3: Solve using the quadratic formula:
  $$x=\frac{-4\pm \sqrt{4^{2}-4\left(-m\right)\left(-3\right)}}{2\left(-m\right)}$$
- step4: Simplify the expression:
  $$x=\frac{-4\pm \sqrt{4^{2}-4\left(-m\right)\left(-3\right)}}{-2m}$$
- step5: Simplify the expression:
  $$x=\frac{-4\pm \sqrt{16-12m}}{-2m}$$
- step6: Simplify the expression:
  $$x=\frac{-4\pm 2\sqrt{4-3m}}{-2m}$$
- step7: Separate into possible cases:
  $$\begin{align}&x=\frac{-4+2\sqrt{4-3m}}{-2m}\&x=\frac{-4-2\sqrt{4-3m}}{-2m}\end{align}$$
- step8: Simplify the expression:
  $$\begin{align}&x=\frac{2-\sqrt{4-3m}}{m}\&x=\frac{-4-2\sqrt{4-3m}}{-2m}\end{align}$$
- step9: Simplify the expression:
  $$\begin{align}&x=\frac{2-\sqrt{4-3m}}{m}\&x=\frac{2+\sqrt{4-3m}}{m}\end{align}$$
The roots of the equation $$4x-3=m*x^{2}$$ are given by the solutions to the quadratic equation $$4x-3=m*x^{2}$$.

Given that -4 is one of the roots, we can substitute $$x=-4$$ into the equation to find the value of $$m$$.

Substituting $$x=-4$$ into the equation $$4x-3=m*x^{2}$$, we get:

$$4(-4)-3=m*(-4)^{2}$$

Simplifying this equation, we have:

$$-16-3=16m$$

$$-19=16m$$

$$m=-\frac{19}{16}$$

Therefore, the value of $$m$$ is $$-\frac{19}{16}$$.

To find the other root, we can substitute $$m=-\frac{19}{16}$$ back into the solutions for $$x$$:

$$x=\frac{2-\sqrt{4-3m}}{m}$$

$$x=\frac{2-\sqrt{4-3(-\frac{19}{16})}}{-\frac{19}{16}}$$

$$x=\frac{2-\sqrt{4+\frac{57}{16}}}{-\frac{19}{16}}$$

$$x=\frac{2-\sqrt{\frac{64+57}{16}}}{-\frac{19}{16}}$$

$$x=\frac{2-\sqrt{\frac{121}{16}}}{-\frac{19}{16}}$$

$$x=\frac{2-\frac{11}{4}}{-\frac{19}{16}}$$

$$x=\frac{\frac{8-11}{4}}{-\frac{19}{16}}$$

$$x=\frac{-\frac{3}{4}}{-\frac{19}{16}}$$

$$x=\frac{3}{4} \times \frac{16}{19}$$

$$x=\frac{12}{19}$$

Therefore, the other root of the equation is $$\frac{12}{19}$$.
$$ m = -\frac{19}{16} $$ and the other root is $$ \frac{12}{19} $$.

2.1. Given \( \frac{4 x-3}{x^{2}}=m \), if -4 is one of the roots of the equation, determine the value of \( m \) and the other root.

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