Answer:
[tex]x=\pm3\text{ or }x=\pm2[/tex]
Step-by-step explanation:
We have the equation:
[tex]x^4-13x^2+36=0[/tex]
Notice that this resembles a quadratic. So, we can use u-substitution to solve for the values of x.
Let [tex]u=x^2[/tex]. Then:
[tex](x^2)^2-13(x^2)+36=0[/tex]
Perform the substitution:
[tex]u^2-13u+36=0[/tex]
We can now factor using -9 and -4:
[tex](u-9)(u-4)=0[/tex]
Zero Product Property:
[tex]u-9=0\text{ or } u-4=0[/tex]
Solve for each case:
[tex]u=9\text{ or } u=4[/tex]
Substitute back u:
[tex]x^2=9\text{ or } x^2=4[/tex]
Take the square root of both sides for both equations. Since we’re taking an even root, we will need plus/minus. Therefore, our possible values of x are:
[tex]x=\pm3\text{ or }x=\pm2[/tex]
