First slide

Theory of equations

Question

The real x satisfying the equation $x^{3} + \frac{1}{x^{3}} + x^{2} + \frac{1}{x^{2}} - 6 (x + \frac{1}{x}) - 7 = 0$ is/are

Moderate

A

$\frac{3 + \sqrt{5}}{2}$
B

$\frac{- 3 - \sqrt{5}}{2}$
C

$\frac{3 - \sqrt{5}}{2}$
D

$\frac{- 3 + \sqrt{5}}{2}$

Solution

Let $x + \frac{1}{x} = t$
Then given equation is
$(t^{3} - 3 t) + (t^{2} - 2) - 6 (t) - 7 = 1$
$\begin{array}{l} or t^{3} + t^{2} - 9 t - 9 = 0 \\ ∴ (t + 1) (t^{2} - 9) = 0 \\ ∴ t = - 1, 3, - 3 \end{array}$
If $x + \frac{1}{x} = - 1$ , then $x^{2} + x + 1 = 0$ . This equation has no real roots.
If $x + \frac{1}{x} = - 3$ , then $x^{2} + 3 x + 1 = 0$ . Roots are $\frac{- 3 \pm \sqrt{5}}{2}$ .
If $x + \frac{1}{x} = 3$ , then $x^{2} - 3 x + 1 = 0$ . Roots are $\frac{3 \pm \sqrt{5}}{2}$ .

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