How do you complete the square to solve $2 x^{2} - 10 x - 20 = 8$ $2x^2 - 10x - 20 = 8$ ?

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How do you complete the square to solve $2 x^{2} - 10 x - 20 = 8$ $2x^2 - 10x - 20 = 8$ ?

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Answer 1 · 2015-05-23T23:19:56

First notice that all the terms are divisible by $2$ $2$ , so let us divide both sides of the equation by $2$ $2$ to get:

$4 = x^{2} - 5 x - 10$ $4=x^2-5x-10$

$= x^{2} - 5 x + \frac{25}{4} - \frac{25}{4} - 10$ $= x^2-5x+25/4-25/4-10$

$= {(x - \frac{5}{2})}^{2} - \frac{25}{4} - \frac{40}{4}$ $= (x-5/2)^2-25/4-40/4$

$= {(x - \frac{5}{2})}^{2} - \frac{65}{4}$ $= (x-5/2)^2 - 65/4$

Add $\frac{65}{4}$ $65/4$ to both ends to get:

${(x - \frac{5}{2})}^{2} = 4 + \frac{65}{4} = \frac{16}{4} + \frac{65}{4} = \frac{81}{4} = {(\frac{9}{2})}^{2}$ $(x-5/2)^2 = 4+65/4 = 16/4+65/4 = 81/4 = (9/2)^2$

So

$(x - \frac{5}{2}) = \pm \frac{9}{2}$ $(x-5/2) = +-9/2$

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

$x = \frac{5}{2} \pm \frac{9}{2}$ $x = 5/2+-9/2$

That is $x = - \frac{4}{2} = - 2$ $x = -4/2 = -2$ or $x = \frac{14}{2} = 7$ $x = 14/2 = 7$ .

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How do you complete the square to solve $2 x^{2} - 10 x - 20 = 8$ $2x^2 - 10x - 20 = 8$ ?

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