How do you complete the square to solve $2 k^{2} + 2 k = 10$ $2k^2 + 2k = 10$ ?

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How do you complete the square to solve $2 k^{2} + 2 k = 10$ $2k^2 + 2k = 10$ ?

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Answer 1 · 2015-06-12T05:59:09

Divide by the coefficient of $k^{2}$ $k^2$ , then add the square of half the coefficient of $k$ $k$ to both sides.

Explanation:

First, divide everything in the equation by two, to make sure that $k^{2}$ $k^2$ has a coefficient of $1$ $1$

$k^{2} + k = 5$ $k^2+k=5$

Next, add ${(\frac{1}{2})}^{2}$ $(1/2)^2$ to both sides of the equation

$k^{2} + k + {(\frac{1}{2})}^{2} = 5 + {(\frac{1}{2})}^{2}$ $k^2 + k + (1/2)^2 = 5 + (1/2)^2$

The left hand side is now a perfect square

${(k + \frac{1}{2})}^{2} = 5 + {(\frac{1}{2})}^{2}$ $(k+1/2)^2 = 5 + (1/2)^2$

Because the right hand side is positive, you can take the $\pm \sqrt{}$ $+-sqrt()$ of both sides

$k + \frac{1}{2} = \pm \sqrt{5 + \frac{1}{4}}$ $k+1/2 = +-sqrt(5 + 1/4$

Simply subtract 1/2 from both sides for your final answer

$k = - \frac{1}{2} \pm \sqrt{5 \frac{1}{4}} = - \frac{1}{2} \pm \sqrt{\frac{21}{4}} = - \frac{1}{2} \pm \frac{\sqrt{21}}{2}$ $k=-1/2+-sqrt(5 1/4)=-1/2+-sqrt(21/4)=-1/2+-sqrt(21)/2$ . Finis!

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How do you complete the square to solve $2 k^{2} + 2 k = 10$ $2k^2 + 2k = 10$ ?

1 Answer

Explanation:

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How do you complete the square to solve $2 k^{2} + 2 k = 10$ $2k^2 + 2k = 10$ ?