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- Feb 14, 2012

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- Feb 14, 2012

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- Feb 14, 2012

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By substituting $x=m+\dfrac{1}{2}$ into the equation yields $8m^4+4m^2+p-\dfrac{3}{2}=0$.

Using the substitution skill one more time (by letting $n=m^2$), we get a quadratic equation in $n$:

$8n^2+4n+p-\dfrac{3}{2}=0$

Solving for $n$ by employing the quadratic formula we get

$n_1=\dfrac{-1+\sqrt{2(2-p)}}{4}$ and $n_2=\dfrac{-1-\sqrt{2(2-p)}}{4}$

If we want real $m$ values, we need $n \ge 0$.

Obviously $n_2=\dfrac{-1-\sqrt{2(2-p)}}{4} <0$, hence it gives only non-real values for $m$ for all real $p$.

Therefore, we can conclude that the equation $8x^4-16x^3+16x^2-8x+p=0$ has at least one non-real root for every real number $p$.

Now, for the second part of the problem, we are asked to find the sum of all the non-real roots of the equation.

For $p \le \dfrac{3}{2}$: | For $p>\dfrac{3}{2}$: |

If we want $n_1=\dfrac{-1+\sqrt{2(2-p)}}{4}>0$, this is true when $p \le\dfrac{3}{2}$. That means, the original equation $8x^4-16x^3+16x^2-8x+p=0$ has 2 real roots and 2 non-real roots and its non-real roots take the form $x=m+\dfrac{1}{2}$, where $m=\sqrt{n}=\sqrt{\text{negative value}}=ai$. The sum of the 2 non-real roots therefore is $\left(ai+\dfrac{1}{2} \right)+\left(-ai+\dfrac{1}{2} \right)=1$. | We will get $n_1=\dfrac{-1+\sqrt{2(2-p)}}{4}<0$. That means, the original equation $8x^4-16x^3+16x^2-8x+p=0$ has all 4 non-real roots and its non-real roots take the form $x=m_1+\dfrac{1}{2}$ and $x=m_2+\dfrac{1}{2}$, where $m_1=\sqrt{n}=\sqrt{\text{negative value}}=ci$ and $m_2=\sqrt{n}=\sqrt{\text{negative value}}=di$ The sum of the 4 non-real roots therefore is $\left(ci+\dfrac{1}{2} \right)+\left(-ci+\dfrac{1}{2} \right)+\left(di+\dfrac{1}{2} \right)+\left(-di+\dfrac{1}{2} \right)=2$. |

Therefore we get:

$\displaystyle \text{the sum of non-real roots}=\begin{cases}1 & \text{for p} \le \dfrac{3}{2} \\2 & p > \dfrac{3}{2}\\ \end{cases}$