What is the absolute value of the difference between the roots of $x^{2} + 6 x + 5 = 0$ ?

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Solution

The quadratic polynomial $x^{2} + 6 x + 5$ can be factorised as follows
$x^{2} + 6 x + 5$
= $x^{2} + 5 x + x + 5$
= $x (x + 5) + 1 (x + 5) 44$
= $(x + 5) (x + 1)$
Therefore the given quadratic equation becomes $(x + 5) (x + 1) = 0$
This gives $x = - 5$ or $x = - 1$
$∴$ the required absolute difference between the roots is $| - 5 - (- 1) | = 4$

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