Question

what will be the solution of this question? with explanation

Answer 1

you can solve it by linear equation

Answer 2

$(x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2$

Let $x - \frac{1}{x}$ be k.

∴ $(x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 = k^2 \\ \\ x^2 + \frac{1}{x^2} = k^2 + 2$

∴ The equation will be like,

$6(k^2 + 2) - k + 2 = 0 \\ \\ = 6k^2 + 12 - k + 2 = 0 \\ \\ = 6k^2 - k + 14 = 0$

Here, $k = \frac{-b + \sqrt{b^2 - 4ac}}{2a} OR \frac{-b - \sqrt{b^2 - 4ac}}{2a}$

$a = 6 \\ b = -1 \\ c = 14 \\ \\ \sqrt{b^2 - 4ac} = \sqrt{(-1)^2 - (4 * 6 * 14)} \\ = \sqrt{1 - 336} = \sqrt{-335}$

It is impossible to find if and only if b² < 4ac or b² - 4ac gives a negative number.

So there isn't any equation like the one in the question.

Hope this may be helpful.

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