Question

the roots of the equation: ^4−^2−12=0 are:a. There are two equal and real roots and two unequal complex rootsb. There are two unequal real roots and two equal complex rootsc. There are two unequal real roots and two unequal complex rootsd. There are three equal real roots and one complex roote. None of the above

Answer 1

We have the next equation:

`x^4-x^2-12=0`

Substituting with:

`\begin{gathered} y=x^2 \\ y^2=(x^2)^2=x^4 \end{gathered}`

we get:

`y^2-y-12=0`

Applying the quadratic formula to the last equation:

`\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)=(1\pm7)/(2) \\ y_1=(1+7)/(2)=4 \\ y_2=(1-7)/(2)=-3 \end{gathered}`

Coming back to the original variable:

`\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}`

And with the other solution:

`\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}`

In conclusion, there are two unequal real roots (x1 and x2) and two unequal complex roots (x3 and x4).