Question

X^-2+4x^-1+3=0 solve by making appropriate substitution

Answer 1

ANSWER

$x = - 1 \: or \: x = - \frac{1}{ 3}$

EXPLANATION

The given equation is:

${x}^{ - 2} + 4 {x}^{ - 1} + 3 = 0$

Recall that:

${a}^{ - m} = \frac{1}{ {a}^{m} }$

$\frac{1}{ {x}^{2} } + \frac{4}{x} + 3 = 0$

Or

${( \frac{1}{x} )}^{2} + 4( \frac{1}{x} ) + 3 = 0$

Let

$u = \frac{1}{x}$

Our equation then becomes:

${u}^{2} + 4u + 3 = 0$

The factors of 3 that add up to 4 are:

${u}^{2} + 3u + u + 3$

$u(u + 3) + 1(u + 3) = 0$

$(u + 1)(u + 3) = 0$

$u + 1 = 0 \: or \: u + 3 = 0$

$u = - 1 \: or \: u = - 3$

This implies that:

$\frac{1}{x} = - 1 \: or \: \frac{1}{x} = - 3$

$x = - 1 \: or \: x = - \frac{1}{ 3}$