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Relation between Roots and Coefficients for Quadratic

The number of...

Question

The number of non integer solutions of the equation ${log}_{x^{2} - 6 x + 8} ({log}_{{2 x}^{2} - 2 x - 8} (x^{2} + 5 x)) = 0$ is/are

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Solution

This equation is equivalent to the system
$⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x^{2} - 6 x + 8 > 0 x^{2} - 6 x + 8 \neq 1 {2 x}^{2} - 2 x - 8 > 0 {2 x}^{2} - 2 x - 8 \neq 1 {log}_{2 x^{2} - 2 x - 8} (x^{2} + 5 x) > 0 {log}_{2 x^{2} - 2 x - 8} (x^{2} - 5 x) = (x^{2} - 6 x + 8)^{0} = 1 \Rightarrow x^{2} + 5 x = {2 x}^{2} - 2 x - 8 \end{matrix}$
Solve the equation of the system
$\Rightarrow ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x < 2 a n d x > 4 x \neq 3 \pm \sqrt{2} x < \frac{1 - \sqrt{17}}{2} a n d x > \frac{1 + \sqrt{17}}{2} x \neq \frac{1 \pm \sqrt{19}}{2} x = - 1, 8 \end{matrix}$
$x = - 1$ , does not satisfy the third relation of this system. Hence $x = 8$ is only root of the original equation.

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