How do you solve using the completing the square method $2 x^{2} + 7 x - 5 = 0$ $2x^2+7x-5=0$ ?

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How do you solve using the completing the square method $2 x^{2} + 7 x - 5 = 0$ $2x^2+7x-5=0$ ?

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Answer 1 · 2018-03-03T11:31:37

$x = (\frac{1}{4}) (3 \sqrt{11} - 7), - (\frac{1}{4}) (3 \sqrt{11} + 7)$ $color(blue)(x = (1/4) (3sqrt11 - 7), -(1/4)(3sqrt11 + 7)$

Explanation:

$2 x 2 + 7 x - 5 = 0$ $2x2 + 7x - 5 = 0$

Dividing by 2, $x 2 + 2 \cdot (\frac{7}{4}) x = \frac{5}{2}$ $x2 + 2 * (7/4) x = 5/2$

Adding ${(\frac{7}{4})}^{2}$ $(7/4)^2$ to both sides,

$x^{2} = (2 \cdot (\frac{7}{4}) x + {(\frac{7}{4})}^{2} = \frac{5}{2} + {(\frac{7}{4})}^{2} = \frac{99}{16}$ $x^2 = (2*(7/4)x + (7/4)^2 = 5/2 + (7/4)^2 = 99/16$

${(x + \frac{7}{4})}^{2} = {(\sqrt{\frac{99}{16}})}^{2}$ $(x+7/4)^2 = (sqrt(99/16))^2$

$x + \frac{7}{4} = \pm \sqrt{\frac{99}{16}}$ $x + 7/4 = +- sqrt(99/16)$

$x = \sqrt{\frac{99}{16}} - \frac{7}{4}, - \sqrt{\frac{99}{16}} - \frac{7}{4}$ $x = sqrt(99/16) - 7/4, -sqrt(99/16 ) - 7/4$

$x = (\frac{1}{4}) (3 \sqrt{11} - 7), - (\frac{1}{4}) (3 \sqrt{11} + 7)$ $x = (1/4) (3sqrt11 - 7), -(1/4)(3sqrt11 + 7)$

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How do you solve using the completing the square method $2 x^{2} + 7 x - 5 = 0$ $2x^2+7x-5=0$ ?

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