How do you solve the equation $x^{2} + 7 x + 15 = 0$ $x^2+7x+15=0$ by completing the square?

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How do you solve the equation $x^{2} + 7 x + 15 = 0$ $x^2+7x+15=0$ by completing the square?

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Answer 1 · 2017-04-05T15:50:45

${(x + \frac{7}{2})}^{2} + \frac{11}{4} = 0$ $(x + 7/2)^2 + 11/4 = 0$

Explanation:

The process of completing the square yields the standard form or
Vertex form $y = a {(x - h)}^{2} + k$ $y = a(x-h)^2 + k$ , where vertex: $(h, k)$ $(h, k)$

Use completing of the square by grouping the $x$ $x$ -terms:
$(x^{2} + 7 x) + 15 = 0$ $(x^2 + 7x) +15 = 0$

Get the squared value by halving the $x$ $x$ -term: $\frac{1}{2} \cdot 7 = \frac{7}{2}$ $1/2 *7 = 7/2$
${(x + \frac{7}{2})}^{2} + 15 - {(\frac{7}{2})}^{2} = 0$ $(x +7/2)^2 +15 - (7/2)^2 = 0$

You need to subtract ${(\frac{7}{2})}^{2} = \frac{49}{4}$ $(7/2)^2 = 49/4$ because it is added when the square is completed:

${(x + \frac{7}{2})}^{2} = (x + \frac{7}{2}) (x + \frac{7}{2}) = x^{2} + 7 x + \frac{49}{4}$ $(x+7/2)^2 = (x+7/2)(x+7/2) = x^2 +7x + 49/4$

From ${(x + \frac{7}{2})}^{2} + 15 - \frac{49}{4} = 0$ $" " (x + 7/2)^2 +15 - 49/4 = 0$

${(x + \frac{7}{2})}^{2} + \frac{60}{4} - \frac{49}{4} = 0$ $" " (x + 7/2)^2 +60/4 - 49/4 = 0$

${(x + \frac{7}{2})}^{2} + \frac{11}{4} = 0$ $(x+7/2)^2 + 11/4 = 0$

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How do you solve the equation $x^{2} + 7 x + 15 = 0$ $x^2+7x+15=0$ by completing the square?

1 Answer

Explanation:

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How do you solve the equation $x^{2} + 7 x + 15 = 0$ $x^2+7x+15=0$ by completing the square?