How do you solve using the completing the square method $x^{2} + 3 x - (\frac{7}{4}) = 0$ $x^2 + 3x - (7/4) = 0$ ?

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How do you solve using the completing the square method $x^{2} + 3 x - (\frac{7}{4}) = 0$ $x^2 + 3x - (7/4) = 0$ ?

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Answer 1 · 2016-05-07T07:53:38

Completing the square allows you to find the maximum or minimum point of a quadratic function. See working below.

Explanation:

$x^{2}$ $x^2$ + 3x - $(\frac{7}{4})$ $(7/4)$ = 0
${(x + (\frac{3}{2}))}^{2}$ $(x + (3/2))^2$ - ${(\frac{3}{2})}^{2}$ $(3/2)^2$ - $(\frac{7}{4})$ $(7/4)$ =0
${(x + \frac{3}{2})}^{2}$ $(x + 3/2)^2$ - 4 =0
Hence the vertex has a coordinate of y= -4 and x = - $\frac{3}{2}$ $3/2$
Maximum point = ( $- \frac{3}{2}$ $-3/2$ , -4)

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How do you solve using the completing the square method $x^{2} + 3 x - (\frac{7}{4}) = 0$ $x^2 + 3x - (7/4) = 0$ ?

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