Let , be the roots of the equation x^2-√2x+2=0.

Question

Let $\alpha ,\beta $ be the roots of the equation ${x}^{2}-\sqrt{2}x+2=0$. Then ${\alpha }^{14}+{\beta }^{14}$ is equal to

Accepted Answer

Find the roots of the quadratic equation:

The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. For the quadratic equation $$x^2 - \sqrt{2}x + 2 = 0$$, we have $$a = 1, b = -\sqrt{2}, c = 2$$. Plugging these into the quadratic formula gives:

$$x = \frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2 - 4(1)(2)}}{2(1)} = \frac{\sqrt{2} \pm \sqrt{2 - 8}}{2} = \frac{\sqrt{2} \pm \sqrt{6}i}{2}.$$

So, we have two roots, $\alpha$ and $\beta$, which are:

$$\alpha = \frac{\sqrt{2} + \sqrt{6}i}{2},$$

$$\beta = \frac{\sqrt{2} - \sqrt{6}i}{2}.$$

Express the roots in exponential form:

We can express complex numbers in the form $$re^{i\theta}$$. For $\alpha$ and $\beta$, we find the magnitude $$r = \sqrt{2}$$ and the arguments $$\theta = \frac{\pi}{3}, -\frac{\pi}{3}$$ respectively. So, we have:

$$\alpha = \sqrt{2}e^{i\frac{\pi}{3}},$$

$$\beta = \sqrt{2}e^{-i\frac{\pi}{3}}.$$

Calculate the 14th power of the roots:

To find $$\alpha^{14}$$ and $$\beta^{14}$$, we use the property of exponents which says that $(a^m)^n = a^{mn}$. So, we have:

$$\alpha^{14} = (\sqrt{2}e^{i\frac{\pi}{3}})^{14} = 2^7e^{i\frac{14\pi}{3}} = 128e^{i\frac{2\pi}{3}},$$

$$\beta^{14} = (\sqrt{2}e^{-i\frac{\pi}{3}})^{14} = 2^7e^{-i\frac{14\pi}{3}} = 128e^{-i\frac{2\pi}{3}}.$$

Add the 14th powers of the roots:

We want to find the real part of $$\alpha^{14} + \beta^{14}$$. To do this, we use the property that $$e^{ix} = \cos(x) + i\sin(x)$$. We have:

$$\alpha^{14} + \beta^{14} = 128e^{i\frac{2\pi}{3}} + 128e^{-i\frac{2\pi}{3}} = 128(2)\cos\left(\frac{2\pi}{3}\right) = -128.$$

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