Find the roots of `x^4+6x^2+14=0 ` . This is a quadratic in `x^2 ` , so we can use any techniques that we know to find the roots.

Using the quadratic formula, if `at^2+bt+c=0 ` then `t=(-b +- sqrt(b^2-4ac))/(2a) `

Here `t=x^2 ` so:

`x^2=(-6 +- sqrt(36-4(1)(14)))/(2) `

Since the discriminant (the expression in the radicand) is negative, there are no real roots.

`x^2=-3 +- sqrt(-20)/2 = -3 +- isqrt(5) `

Thus `x=+-sqrt(-3+-isqrt(5)) `

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The roots are `x=sqrt(-3+isqrt(5)),sqrt(-3-isqrt(5)),-sqrt(-3+isqrt(5)),-sqrt(-3-isqrt(5)) `

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Since the polynomial is of 4th degree, we know by the fundamental theorem of algebra that there are 4 roots in the complex numbers.

We can tell that none of the roots are real since (1) the discriminant is negative or (2) the graph is always above the x-axis. We could also apply the rule of signs.

The graph: