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Definite Integral as Limit of Sum

Solve for x...

Question

Solve for $x$ .
$| x^{2} - 5 x | > | x^{2} | - | 5 x |$

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Solution

At $x = 0, x = 5, | x^{2} - 5 x | = | x^{2} | - | 5 x | = 0$
For $x \in (0, 5), | x^{2} | - | 5 x | < 0$ but, $| x^{2} - 5 x | > 0$ .
For $x > 5, | x^{2} - 5 x | = | x^{2} | - | 5 x |$ as all terms are positive.
For $x < 0, | x^{2} | - | 5 x | < 0$ but, $| x^{2} - 5 x | > 0$ .
Hence, $| x^{2} - 5 x | > | x^{2} | - | 5 x | \forall x \in (- \infty, 0) \cup (0, 5)$

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