How do you solve using the completing the square method $2 x^{2} = 2 x + 15$ $2x^2 = 2x + 15$ ?

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How do you solve using the completing the square method $2 x^{2} = 2 x + 15$ $2x^2 = 2x + 15$ ?

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Answer 1 · 2017-04-29T11:34:33

Solution : $x = \frac{1}{2} (1 + \sqrt{31}) and x = \frac{1}{2} (1 - \sqrt{31})$ $x= 1/2(1 + sqrt(31)) and x= 1/2(1 - sqrt(31))$

Explanation:

$2 x^{2} = 2 x + 15 or 2 x^{2} - 2 x = 15 or 2 (x^{2} - x) = 15 or (x^{2} - x) = \frac{15}{2}$ $2x^2 =2x+15 or 2x^2 -2x= 15 or 2(x^2-x) =15 or (x^2-x) = 15/2$ or

$(x^{2} - x + {(\frac{1}{2})}^{2}) - \frac{1}{4} = \frac{15}{2} or {(x - \frac{1}{2})}^{2} = \frac{15}{2} + \frac{1}{4} or {(x - \frac{1}{2})}^{2} = \frac{31}{4}$ $(x^2-x +(1/2)^2) -1/4= 15/2 or (x-1/2)^2 = 15/2+1/4 or (x-1/2)^2 = 31/4$ or
$(x - \frac{1}{2}) = \pm \sqrt{\frac{31}{4}} or x = \frac{1}{2} \pm \frac{\sqrt{31}}{2} or x = \frac{1}{2} (1 \pm \sqrt{31})$ $(x-1/2)= +-sqrt(31/4) or x = 1/2+- sqrt(31)/2 or x = 1/2(1 +- sqrt(31))$

Solution : $x = \frac{1}{2} (1 + \sqrt{31}) and x = \frac{1}{2} (1 - \sqrt{31})$ $x= 1/2(1 + sqrt(31)) and x= 1/2(1 - sqrt(31))$ [Ans]

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How do you solve using the completing the square method $2 x^{2} = 2 x + 15$ $2x^2 = 2x + 15$ ?

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How do you solve using the completing the square method $2 x^{2} = 2 x + 15$ $2x^2 = 2x + 15$ ?