Respuesta :
We want to factor the equation
[tex]81x^4-64x^2=0[/tex]But this is a difference of 2 squares problem, since
[tex]81x^4=(9x^2)^2[/tex]and
[tex]64x^2=(8x)^2[/tex]Thus,
[tex]\begin{gathered} 81x^4-64x^2=(9x^2)^2-(8x)^2 \\ =(9x^2-8x)(9x^2+8x)=0 \end{gathered}[/tex]This can further be factorized as,
[tex](x)(x)(9x-8)(9x+8)[/tex]This is the solution to STEP 1, as you can see, there are 4 factors.
2. We want to get the solutions from this factored form, i.e the values of x that makes
[tex](x)(x)(9x-8)(9x+8)=0[/tex]We see that this happens, when x = 0 twice, and when
[tex]\begin{gathered} 9x-8=0 \\ 9x=8 \\ x=\frac{8}{9} \end{gathered}[/tex]also when,
[tex]\begin{gathered} 9x+8=0 \\ 9x=-8 \\ x=-\frac{8}{9} \end{gathered}[/tex]Therefore the solutions are
[tex]0,0,\frac{8}{9},-\frac{8}{9}[/tex]This is the answer to Part B
