Question 1061504

Solve the following equation by first doing a substitution to make it a quadratic. 

3x^-2 + x^-1 -24=0
<pre>{{{3x^(- 2) + x^(- 1) - 24 = 0}}}
The SIMPLE substitution: Let {{{matrix(1,3, a, be, x^(- 1))}}}
Then {{{matrix(1,3, 3x^(- 2), becomes, 3a^2)}}}, and {{{3x^(- 2) + x^(- 1) - 24 = 0}}} becomes: {{{3a^2 + a - 24 = 0}}}
{{{3a^2 + 9a - 8a - 24 = 0}}} ------ Repalacing a, with 9a - 8a
3a(a + 3) - 8(a + 3) = 0
(a + 3)(3a - 8) = 0
{{{matrix(1,7, a, "=", - 3, or, a, "=", 8/3)}}} 

{{{x^(- 1) = - 3}}} ------- Substituting back {{{matrix(1,5, a, or, - 3, for, x^(- 1))}}}
{{{x^((- 1) * (- 1)) = (- 3)^(- 1)}}} ------ Raising each side to the - 1 power
{{{highlight_green(x = - 1/3)}}} 
Substitute this value for x ({{{- 1/3}}}) into the original equation. If this value satisfies the original equation, then {{{x = - 1/3}}} is a solution to the equation. 
This value for x DOES SATISFY the equation.

{{{x^(- 1) = (- 8)/3}}} ------- Substituting back {{{matrix(1,5, a, or, 8/3, for, x^(- 1))}}}
{{{x^((- 1) * (- 1)) = (8/3)^(- 1)}}} ------ Raising each side to the - 1 power
{{{highlight_green(x = 3/8)}}} 
Substitute this value for x ({{{3/8}}}) into the original equation. If this value satisfies the original equation, then {{{x = 3/8}}} is a solution to the equation.
This value for x DOES SATISFY the equation.