Chapter 1.6, Problem 50E

To determine

To calculate: The solution of the equation $4x^2{\left(x-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6x{\left(x-1\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=0$.

The solutions of the given equation are 0,1, and $\frac{3}{5}$.

Calculation:

Consider the provided equation,

$4x^2{\left(x-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6x{\left(x-1\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=0$

Take the common factor out.

$\begin{array}{c}2x{\left(x-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left[2x+3\left(x-1\right)\right]=0\\ 2x{\left(x-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(5x-3\right)=0\end{array}$

Put the first factor equal to zero.

$x=0$

Put the second factor equal to zero.

$\begin{array}{c}{\left(x-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=0\\ x=1\end{array}$

Put the third factor equal to zero.

$\begin{array}{c}5x-3=0\\ x=\frac{3}{5}\end{array}$

Check:

Put $x=0$ in the equation $4x^2{\left(x-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6x{\left(x-1\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=0$.

$\begin{array}{r}4{\left(0\right)}^2{\left(\left(0\right)-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6\left(0\right){\left(\left(0\right)-1\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\stackrel{?}{=}0\\ 4\left(0\right)+6\left(0\right)\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Put $x=1$ in the equation $4x^2{\left(x-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6x{\left(x-1\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=0$.

$\begin{array}{r}4{\left(1\right)}^2{\left(\left(1\right)-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6\left(1\right){\left(\left(1\right)-1\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\stackrel{?}{=}0\\ 4\left(0\right)+6\left(0\right)\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Put $x=\frac{3}{5}$ in the equation $4x^2{\left(x-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6x{\left(x-1\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=0$.

$\begin{array}{r}4{\left(\frac{3}{5}\right)}^2{\left(\left(\frac{3}{5}\right)-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6\left(\frac{3}{5}\right){\left(\left(\frac{3}{5}\right)-1\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\stackrel{?}{=}0\\ 4{\left(\frac{3}{5}\right)}^2{\left(\frac{-2}{5}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6\left(\frac{3}{5}\right){\left(\frac{-2}{5}\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\stackrel{?}{=}0\\ 4\left(0\right)+6\left(0\right)\stackrel{?}{=}0\\ 0=0\end{array}$

Hence, the solution of the given equation is $x=0,1,\frac{3}{5}$.