Let , be the roots of the equation x^2-ax-b=0 with Im( )

Question

Let $\alpha ,\beta $ be the roots of the equation ${x}^{2}-ax-b=0$ with $Im(\alpha )<Im(\beta )$. Let ${P}_{n}={\alpha }^{n}-{\beta }^{n}$. If ${P}_{3}=-5\sqrt{7}i,{P}_{4}=-3\sqrt{7}i,{P}_{5}=11\sqrt{7}i$ and ${P}_{6}=45\sqrt{7}i$, then $|{\alpha }^{4}+{\beta }^{4}|$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 31

Accepted Answer

We begin with the equations for the roots:

$\alpha +\beta =a$

$\alpha \beta =-b$

Given:

${P}_{6}={aP}_{5}+{bP}_{4}$

${P}_{5}={aP}_{4}+{bP}_{3}$

Using the given values:

For ${P}_{6}$:

$45\sqrt{7}i=a	imes 11\sqrt{7}i+b(-3\sqrt{7})i$

Simplifying, we obtain:

$45=11a-3b\,$

For ${P}_{5}$:

$11\sqrt{7}i=a(-3\sqrt{7}i)+b(-5\sqrt{7}i)$

Simplifying, we obtain:

$11=-3a-5b\,$

Solving these linear equations, we find:

$a=3$

$b=-4$

Now, we calculate $|{\alpha }^{4}+{\beta }^{4}|$ using the relation:

$|{\alpha }^{4}+{\beta }^{4}|=\sqrt{({\alpha }^{4}-{\beta }^{4}{)}^{2}+4({\alpha }^{4}{\beta }^{4})}$

From $b=-4$, we know:

$\alpha \beta =-b=4\, \Rightarrow {\alpha }^{4}{\beta }^{4}=(\alpha \beta {)}^{4}={4}^{4}=256$

Substitute into the relation:

$|{\alpha }^{4}+{\beta }^{4}|=\sqrt{(-63)+1024}$

$=\sqrt{961}=31$

Thus, $|{\alpha }^{4}+{\beta }^{4}|$ is equal to 31.

Solutions · 1