Respuesta :
The given expression is :
[tex]y=x^4-7x^2+12[/tex]To find the roots :
[tex]\begin{gathered} \text{ Let u=x}^2\text{ then } \\ u^2=x^4 \end{gathered}[/tex]So, the equation will be :
[tex]x^4-7x^2+12=u^2-7u+12[/tex]Simplify the u by factorization method :
[tex]\begin{gathered} u^2-7u+12=u^2-3u-4u+12 \\ u^2-7u+12=u(u-3)-4(u-3) \\ u^2-7u+12=(u-4)(u-3)_{} \end{gathered}[/tex]as : u = x² , substitute these value back :
[tex]\begin{gathered} u^2-7u+12=(u-4)(u-3) \\ u^2-7u+12=(x^2-4)(x^2-3) \end{gathered}[/tex]Apply the quadratic expression : (a² - b²) = (a - b)(a + b)
[tex]\begin{gathered} u^2-7u+12=\mleft(x^2-4\mright)\mleft(x^2-3\mright) \\ x^4^{}-7x^2+12=(x^2-2^2_{})(x^2-\sqrt[]{3}^2) \\ x^4-7x^2+12=(x-2)(x+2)(x-\sqrt[]{3})(x+\sqrt[]{3}) \end{gathered}[/tex][tex]\text{ So, the factors are : }=(x-2)(x+2)(x-\sqrt[]{3})(x+\sqrt[]{3})[/tex]
For the roots, equate each factor with zero:
[tex]\begin{gathered} (x-2)=0\text{ }\Rightarrow x=2 \\ (x+2)=0\Rightarrow x=-2 \\ (x-\sqrt[]{3})=0\Rightarrow x=\sqrt[]{3} \\ (x+\sqrt[]{3})=0\Rightarrow x=-\sqrt[]{3} \end{gathered}[/tex]So, the roots are :
[tex]x=2,\text{ -2, }\sqrt[]{3},-\sqrt[]{3}[/tex]Answer :
[tex]x=2,\text{ -2, }\sqrt[]{3},-\sqrt[]{3}[/tex]
