Question

Solve the following equation:
$1 + 8{x^{\dfrac{6}{5}}} + 9\sqrt[5]{{{x^3}}} = 0$

Answer 1

Hint: Here we substitute the terms with a variable so that it is in the form of a quadratic equation and then solve the given equation.

Complete step-by-step answer:
Now we have to solve $1 + 8{x^{\dfrac{6}{5}}} + 9\sqrt[5]{{{x^3}}} = 0$
This can be written as
$1 + 8{\left( {{x^{\dfrac{6}{5}}}} \right)^2} + 9{x^{^{\dfrac{3}{5}}}} = 0$
Now for further simplification put ${x^{\dfrac{3}{5}}} = p$ we will have
$1 + 8{\left( p \right)^2} + 9p = 0$ or $8{p^2} + 9p + 1 = 0$
Now lets factorize this up
$8{p^2} + 8p + p + 1 = 0$
This can be written as
$8p\left( {p + 1} \right) + \left( {p + 1} \right) = 0$
$\left( {8p + 1} \right)\left( {p + 1} \right) = 0$
So we are getting two values of p that is
$p = - 1, - \dfrac{1}{8}$
Now we have assumed that ${x^{\dfrac{3}{5}}} = p$
So our $x = {\left( { - 1} \right)^{\dfrac{5}{3}}},{\left( {\dfrac{{ - 1}}{8}} \right)^{\dfrac{5}{3}}}$
Hence our $x = - 1,{\left( {{{\left( {\dfrac{{ - 1}}{2}} \right)}^3}} \right)^{\dfrac{5}{3}}}$
 $x = - 1,{\left( {\dfrac{{ - 1}}{2}} \right)^5}$
$x = - 1,\dfrac{{ - 1}}{{32}}$

Note: Whenever we need to solve such equation try and simplify by substituting the terms with some other variable in order to obtain a quadratic form which can be solved to find the solution of equation