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

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Answer 1 · 2018-04-12T15:17:12

$x = - 1.5 \pm 2$ $x = -1.5+-2$

$x^{2} + 3 x - \frac{7}{4} = 0$ $x^2+3x−7/4=0$

${(x + \frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} - \frac{7}{4} = 0$ $(x+3/2)^2 - (3/2)^2 - 7/4 = 0$

${(x + \frac{3}{2})}^{2} - \frac{9}{4} - \frac{7}{4} = 0$ $(x+3/2)^2 - 9/4 - 7/4 = 0$

${(x + \frac{3}{2})}^{2} - \frac{16}{4} = 0$ $(x+3/2)^2 - 16/4 = 0$

${(x + \frac{3}{2})}^{2} = 4$ $(x+3/2)^2 = 4$

$x + \frac{3}{2} = \sqrt{4}$ $x+3/2 = sqrt4$

$x = - \frac{3}{2} \pm 2$ $color(red)(x= -3/2+-2)$

$x = 0.5 or x = - 3.5$ $color(red)(x=0.5 or x = -3.5)$

How do you solve x^2 + 3x - (7/4) = 0x2+3x−(74)=0x^2 + 3x - (7/4) = 0 by completing the square?