Value of expression $(\alpha^{110}+\beta^{110})-(\alpha^{98}+\beta^{98})$ in given quadratic equation

Question

If $\alpha,\beta$ are the roots of the equation $x^2+3^{\frac{1}{4}}x+3^{\frac{1}{2}}=0$. Then value of $\displaystyle \alpha^{98}(\alpha^{12}-1)+\beta^{98}(\beta^{12}-1)$ is

From $\displaystyle x^2+3^{\frac{1}{4}}x+3^{\frac{1}{2}}=0$,we have

$\displaystyle \alpha+\beta=-3^{\frac{1}{4}}$ and $\displaystyle \alpha\beta=\sqrt{3}$

Now we have to find value of $\displaystyle (\alpha^{110}+\beta^{110})-(\alpha^{98}+\beta^{98})$

I did not understand how can I write above expression into sum and product of roots

Please have a look

Answer 1

Let $\omega$ and $\omega^2$ be the roots of $x^2 + x + 1 = 0$. Then the roots of the equation $x^2 + 3^{1/4}x + 3^{1/2} = 0$ are $\alpha = 3^{1/4}\omega$ and $\beta = 3^{1/4}\omega^2$.

Using the fact that $\omega^3 = 1$, we have, $\alpha^{12} - 1 = 3^3 \omega^{12} - 1 = 27 - 1 = 26$. Similarly, we can also conclude $\beta^{12} - 1 = 26$, $\alpha^{98} = 3^{24.5} \omega^2$ and $\beta^{98} = 3^{24.5} \omega$. Lastly, by using the relation $\omega + \omega^2 = -1$, we can conclude that \begin{align*} \alpha^{98}(\alpha^{12} - 1) + \beta^{98}(\beta^{12} - 1) = -26 \cdot 3^{24} \cdot \sqrt{3}. \end{align*}

Answer 2

If $\alpha,\beta$ are the roots of the equation $x^2+3^\left({\frac{1}{4}}\right)x+3^\left({\frac{1}{2}}\right)=0$. Then value of $\displaystyle \alpha^{98}(\alpha^{12}-1)+\beta^{98}(\beta^{12}-1)$ is ?

Alternative approach, less elegant than the approach of sudeep5221, but representing a more generic attack. An additional shortcoming of this response is that it is unclear how appropriate the use of (elementary) Complex Analysis is, given the posting's algebra-precalculus tag.

Personally, I was never exposed to Complex Analysis until I was in a Calculus class. On the other hand, this particular problem cries for Complex Analysis.

You can tell at a glance, since $~\displaystyle \left[ ~3^\left({\frac{1}{4}}\right) ~\right]^2 = \left[ ~3^\left({\frac{1}{2}}\right) ~\right],~$ that $~\displaystyle \left[ ~3^\left({\frac{1}{4}}\right) ~\right]^2 - 4\left[ ~3^\left({\frac{1}{2}}\right) ~\right]~ < 0. ~$

Therefore, you know that the roots will be complex, of form $x + iy, x - iy.$

Letting the expression $~e^{i\theta}~$ denote $~\cos(\theta) + i\sin(\theta),~$ you have that for any $\displaystyle ~n \in \Bbb{Z^+}, r \in \Bbb{R^+}, ~\left[r \times e^{i\theta}\right]^n = r^n \times e^{in\theta}.$

Further, it is immediately clear that the two roots will be Complex Conjugates of each other. That is, the roots will have form $~re^{i\theta}, re^{-i\theta}.$

So, this response will derive $~r~$ and $~\theta~$, by solving for the roots' values expressed as $~x \pm iy.$

Then, the Math will be routine. On a side note, given the nature of the problem, I suspect that, despite the algebra-precalculus tag, that my approach will be similar to what the problem composer intended.

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For ease of notation, let $~A~$ denote $~\displaystyle 3^{\left(\frac{1}{4}\right)}.$

Applying the formula for the quadratic equation, the roots will be $x \pm iy,$ where:

• $~\displaystyle x = \frac{1}{2} \times -A = \frac{-A}{2}.$

• $~\displaystyle iy = \frac{1}{2} \times \sqrt{A^2 - 4A^2} = i \times \frac{A}{2} \times \sqrt{3} \implies y = \frac{A\sqrt{3}}{2}.$

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The next step is to compute

$$r = \sqrt{x^2 + y^2} = \sqrt{\frac{A^2}{4} + \frac{3A^2}{4}} = A.$$

Then, the roots $x \pm iy$ can be expressed as

$$A \times \left[ ~\frac{-A}{2A} \pm i \frac{A\sqrt{3}}{2A} ~\right] = A \times \left[\frac{-1}{2} \pm i\frac{\sqrt{3}}{2}\right].$$

Note that

$$\left[ ~\frac{-1}{2}\right]^2 + \left[\frac{\sqrt{3}}{2}\right]^2 = 1.$$

So, the roots have been successfully converted into the form

$$re^{i\theta}, re^{-i\theta} ~: ~r = A = 3^{\left(\frac{1}{4}\right)}, ~\theta = 120^\circ.$$

For simplicity, I will alternatively express $\theta$ as $\pi/3$.

Then, the two roots can be expressed as

$$~\alpha = Ae^{i\pi/3}, ~\beta = Ae^{-i\pi/3}.$$

So,

• $\displaystyle \alpha^{12} = A^{(12)}e^{i4\pi} = A^{12} \times 1 = 3^3 \implies $
  $\alpha^{12} - 1 = 26.$

• $\displaystyle \beta^{12} = A^{(12)}e^{-i4\pi} = A^{12} \times 1 = 3^3 \implies $
  $\beta^{12} - 1 = 26.$
  

• $\displaystyle \alpha^{98} = A^{(98)}e^{i\frac{98\pi}{3}} = 3^{\frac{98}{4}} \times \left[\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right]$
  
  $\displaystyle = 3^{24} \times \sqrt{3} \times \left[ \frac{-1}{2} + i \frac{-\sqrt{3}}{2}\right].$

• $\displaystyle \beta^{98} = A^{(98)}e^{i\frac{-98\pi}{3}} = 3^{\frac{98}{4}} \times \left[\cos\left(\frac{-2\pi}{3}\right) + i\sin\left(\frac{-2\pi}{3}\right)\right]$
  
  $\displaystyle = 3^{24} \times \sqrt{3} \times \left[\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right]$
  
  $\displaystyle = 3^{24} \times \sqrt{3} \times \left[ \frac{-1}{2} + i \frac{\sqrt{3}}{2}\right].$
  
  

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Since $~\displaystyle \alpha^{12}-1 = 26 = \beta^{12} - 1,$
you have that the overall expression has a common factor of $26$.

Further, the overall expression also has a common factor of $\displaystyle 3^{24} \times \sqrt{3}$.

Therefore, the overall expression equals

$$26 \times 3^{24} \times \sqrt{3} \\ \times \left\{ ~\left[ \frac{-1}{2} + i \frac{-\sqrt{3}}{2}\right] + \left[ \frac{-1}{2} + i \frac{\sqrt{3}}{2}\right] ~\right\}$$

$$= 26 \times 3^{24} \times \sqrt{3} \times \{ -1 \}.$$

Answer 3

We can automate the process. Consider what happens if we multiply

$$(\alpha \beta)^k (\alpha + \beta)^{n-2k} =\\ (\alpha \beta)^k \left( \binom{n}{0}\alpha^{n-2k}\beta^0 + \binom{n}{1}\alpha^{n-2k}\beta^1 + \dots \right)= \\ (\alpha \beta)^k \sum_{j=0}^{n-2k}{ \binom{n-2k}{j} \alpha^{n-2k-j}\beta^j }= \\ \sum_{j=0}^{n-2k}{ \binom{n-2k}{j} \alpha^{n-k-j}\beta^{j+k} }$$

It's easy to arrange this as a matrix:

$$\begin{bmatrix} \begin{array}{ccccccc|c} \binom{n}{0} & \binom{n}{1} & \binom{n}{2} & \dots & \binom{n}{n-2} & \binom{n}{n-1} & \binom{n}{n} & (\alpha \beta)^0(\alpha + \beta)^n \\ 0 & \binom{n-2}{0} & \binom{n-2}{1} & \dots & \binom{n-2}{n-1} & \binom{n-2}{n-2} & 0 & (\alpha \beta)^1(\alpha + \beta)^{n-2} \\ 0 & 0 & \binom{n-4}{0} & \dots & \binom{n-4}{n-4} & 0 & 0 & (\alpha \beta)^2(\alpha + \beta)^{n-4} \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots & \vdots \\ \end{array} \end{bmatrix}$$

We must drop the right half of the matrix so that we get something to evaluate:

$$\begin{bmatrix} \begin{array}{ccccc|c} \binom{n}{0} & \binom{n}{1} & \binom{n}{2} & \dots & \binom{n}{n} & (\alpha \beta)^0(\alpha + \beta)^n \\ 0 & \binom{n-2}{0} & \binom{n-2}{1} & \dots & \binom{n-2}{n-2} & (\alpha \beta)^1(\alpha + \beta)^{n-2} \\ 0 & 0 & \binom{n-4}{0} & \dots & \binom{n-4}{n-4} & (\alpha \beta)^2(\alpha + \beta)^{n-4} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 1 & (\alpha \beta)^n (\alpha + \beta)^{0} \end{array} \end{bmatrix}$$

Solving the matrix for the value that corresponds to the leftmost column, we obtain $(a+b)^n$.

To save time, simply factor the powers that we need:

$$\alpha^{110} = \alpha^{2 \cdot 5 \cdot 11}$$ $$\alpha^{98} = \alpha^{2 \cdot 7 \cdot 7}$$

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We can easily find $(\alpha^2 + \beta^2) = (\alpha+\beta)^2 - 2(\alpha \beta) = (-3^{1/4})^2 - 2\sqrt{3} = -\sqrt{3}$

Now, lets take find $(\alpha^{2 \cdot 5} + \beta^{2 \cdot 5})$

$$\begin{bmatrix} \begin{array}{ccc|c} \binom{5}{0} & \binom{5}{1} & \binom{5}{2} & (\alpha^2 \beta^2)^0(\alpha^2 + \beta^2)^5 \\ 0 & \binom{3}{0} & \binom{3}{1} & (\alpha^2 \beta^2)^1(\alpha^2 + \beta^2)^{3} \\ 0 & 0 & \binom{1}{0} & (\alpha^2 \beta^2)^2(\alpha^2 + \beta^2)^{1} \\ \end{array} \end{bmatrix}$$

We find $(\alpha^{10} + \beta^{10}) = - 9\sqrt{3} $.

Next we will find $(\alpha^{10 \cdot 11} + \beta^{10 \cdot 11})$.

$$\begin{bmatrix} \begin{array}{cccccc|c} \binom{11}{0} & \binom{11}{1} & \binom{11}{2} & \binom{11}{3} & \binom{11}{4} & \binom{11}{5} & (\alpha \beta)^{10 \cdot 0}(\alpha^{10} + \beta^{10})^{11} \\ 0 & \binom{9}{0} & \binom{9}{1} & \binom{9}{2} & \binom{9}{3} & \binom{9}{4} & (\alpha \beta)^{10 \cdot 1}(\alpha^{10} + \beta^{10})^{9} \\ 0 & 0 & \binom{7}{0} & \binom{7}{1} & \binom{7}{2} & \binom{7}{3} & (\alpha \beta)^{10 \cdot 2}(\alpha^{10} + \beta^{10})^{7} \\ 0 & 0 & 0 & \binom{5}{0} & \binom{5}{1} & \binom{5}{2} & (\alpha \beta)^{10 \cdot 3}(\alpha^{10} + \beta^{10})^{5} \\ 0 & 0 & 0 & 0 & \binom{3}{0} & \binom{3}{1} & (\alpha \beta)^{10 \cdot 4}(\alpha^{10} + \beta^{10})^{3} \\ 0 & 0 & 0 & 0 & 0 & \binom{1}{0} & (\alpha \beta)^{10 \cdot 5}(\alpha^{10} + \beta^{10})^{1} \\ \end{array} \end{bmatrix}$$

We find $(\alpha^{110} + \beta^{110}) = -7625597484987 \sqrt{3}$.

Similarly, we find $(\alpha^{98} + \beta^{98}) = -282429536481 \sqrt{3}$.

Finally, we arrive at $(\alpha^{110} + \beta^{110}) - (\alpha^{98} + \beta^{98}) = -7343167948506 \sqrt{3}$, which agrees with the other answers.