Chapter 4.7, Problem 31WE

To determine

To find all double roots.

Answer to Problem 31WE

The values of double roots are $0,1,-1,3\text{ and }-3$

Explanation of Solution

Given information:

  $t^6+9t^2=10t^4$

Calculation:

The double rootscan be calculated as

  $\begin{array}{l}{t}^6+9t^2=10t^4\\ t^6+9t^2-10t^4\text{=0 }\left[\text{make one side equal to 0}\right]\\ t^2\left(t^4+9-10t^2\right)=0\\ t^2\left[t^4+9-\left(t^2+9t^2\right)\right]=0\left[\text{write 10}{t}^2=\left(t^2+9t^2\right)\right]\\ t^2\left(t^4+9-t^2-9t^2\right)=0\\ t^2\left[t^2\left(t^2-1\right)-9\left(t^2-1\right)\right]=0\left[\begin{array}{l}\text{the first and third term has a common factor,}{t}^2\text{ and}\\ \text{the second and fourth term has a common factor,9}\end{array}\right]\\ t^2\left(t^2-1\right)\left(t^2-9\right)=0\\ t^2\left[{\left(t\right)}^2-{\left(1\right)}^2\right]\left[{\left(t\right)}^2-{\left(3\right)}^2\right]=0\left[\text{difference of squares}\right]\\ t^2\left(t+1\right)\left(t-1\right)\left(t+3\right)\left(t-3\right)=0\end{array}$

Therefore,

  $\begin{array}{l}{t}^2\text{=0 or }\left(t+1\right)\text{=0 or }\left(t-1\right)=0\text{ or }\left(t+3\right)\text{=0 or }\left(t-3\right)\text{=0 }\left[\text{use the zero-product property}\right]\\ t=0\text{ or }t\text{=}-\text{1 or }t=1\text{ or }t=-3\text{ or }t=3\end{array}$

Hence,

The values of double roots are $0,1,-1,3\text{ and }-3$