QJEEM23Ma0169: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0169 Solution: Let be the roots of the equation . Then is equal to

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:

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