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Sagot :
We have the next equation:
[tex]x^4-x^2-12=0[/tex]Substituting with:
[tex]\begin{gathered} y=x^2 \\ y^2=(x^2)^2=x^4 \end{gathered}[/tex]we get:
[tex]y^2-y-12=0[/tex]Applying the quadratic formula to the last equation:
[tex]\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-(-1)\pm\sqrt[]{(-1)^2-4\cdot1\cdot(-12)}}{2\cdot1} \\ y_{1,2}=\frac{1\pm7}{2} \\ y_1=\frac{1+7}{2}=4 \\ y_2=\frac{1-7}{2}=-3 \end{gathered}[/tex]Coming back to the original variable:
[tex]\begin{gathered} y_{}=x^2 \\ 4=x^2 \\ \sqrt[]{4}=x \\ \text{This equation has two solutions:} \\ 2=x_1 \\ -2=x_2 \end{gathered}[/tex]And with the other solution:
[tex]\begin{gathered} y=x^2 \\ -3=x^2 \\ \sqrt[]{-3}=x \\ \sqrt[]{3\cdot(-1)}=x \\ \sqrt[]{3}\cdot\sqrt[]{-1}=x \\ \sqrt[]{3}i=x \\ \text{This equation has two solutions:} \\ 0+\sqrt[]{3}i=x_3 \\ 0-\sqrt[]{3}i=x_4 \end{gathered}[/tex]In conclusion, there are two unequal real roots (x1 and x2) and two unequal complex roots (x3 and x4).
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