How do you solve by completing the square for $x^{2} - 5 x = 9$ $x^2 - 5x = 9$ ?

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How do you solve by completing the square for $x^{2} - 5 x = 9$ $x^2 - 5x = 9$ ?

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Answer 1 · 2015-06-14T20:06:39

${(x - \frac{5}{2})}^{2} = x^{2} - 5 x + \frac{25}{4} = 9 + \frac{25}{4} = \frac{61}{4}$ $(x-5/2)^2 = x^2-5x+25/4 = 9+25/4 = 61/4$

Hence: $x = \frac{5}{2} \pm \frac{\sqrt{61}}{2}$ $x = 5/2+-sqrt(61)/2$

Explanation:

${(x - \frac{5}{2})}^{2}$ $(x-5/2)^2$

$= x^{2} - 5 x + \frac{25}{4}$ $= x^2-5x+25/4$

$= 9 + \frac{25}{4}$ $= 9+25/4$

$= \frac{36}{4} + \frac{25}{4}$ $= 36/4+25/4$

$= \frac{36 + 25}{4}$ $= (36+25)/4$

$= \frac{61}{4}$ $= 61/4$

So:

$x - \frac{5}{2} = \pm \sqrt{\frac{61}{4}} = \pm \frac{\sqrt{61}}{\sqrt{4}} = \pm \frac{\sqrt{61}}{2}$ $x - 5/2 = +-sqrt(61/4) = +-sqrt(61)/sqrt(4) = +-sqrt(61)/2$

Add $\frac{5}{2}$ $5/2$ to both ends to get:

$x = \frac{5}{2} \pm \frac{\sqrt{61}}{2}$ $x = 5/2+-sqrt(61)/2$

In general,

$a x^{2} + b x + c = a {(x + \frac{b}{2 a})}^{2} + (c - \frac{b^{2}}{4 a})$ $ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))$

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How do you solve by completing the square for $x^{2} - 5 x = 9$ $x^2 - 5x = 9$ ?

1 Answer

Explanation:

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How do you solve by completing the square for $x^{2} - 5 x = 9$ $x^2 - 5x = 9$ ?