Chapter 4.7, Problem 44WE

To determine

To find the multiple roots.

Answer to Problem 44WE

The values of multiple rootsare $\text{1,}-1,2\text{ and }-2$

Explanation of Solution

Given information:

  ${\left(x^2-3\right)}^2+\left(x^2-3\right)-2=0$

Calculation:

The given equation already has 0 as one side,and the other side is already factored. Simply set each factored equal to 0 and solve.

Thedouble rootscan be calculated as

  $\begin{array}{l}{\left(x^2-3\right)}^2+\left(x^2-3\right)-2=0\\ \left[{\left(x^2\right)}^2-2\left(x^2\right)\left(3\right)+{\left(3\right)}^2\right]+\left(x^2-3\right)-2=0\left[\text{perfect square trinomial}\right]\\ \left(x^4-6x^2+9\right)+x^2-3-2=0\\ x^4-6x+9+x^2-5=0\left[\text{simplify}\right]\\ x^4-5x^2+4=0\\ x^4-\left(4x^2+x^2\right)+4=0\\ x^4-4x^2-x^2+4=0\\ x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\left[\begin{array}{l}\text{the first and third term has a common factor,}{x}^2\text{ and}\\ \text{the second and fourth has a common factor,4 }\end{array}\right]\\ \left(x^2-1\right)\left(x^2-4\right)=0\\ \left[{\left(x\right)}^2-{\left(1\right)}^2\right]\left[{\left(x\right)}^2-{\left(2\right)}^2\right]=0\left[\text{difference of squares}\right]\\ \left[\left(x+1\right)\left(x-1\right)\right]\left[\left(x+2\right)\left(x-2\right)\right]=0\end{array}$

Therefore,

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

Hence,

The values of multiple rootsare $\text{1,}-1,2\text{ and }-2$