How do you solve $3 s^{2} - 26 x + 2 = 5 s^{2} + 1$ $3s^2-26x+2=5s^2+1$ by completing the square?

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How do you solve $3 s^{2} - 26 x + 2 = 5 s^{2} + 1$ $3s^2-26x+2=5s^2+1$ by completing the square?

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Answer 1 · 2018-03-03T13:32:09

$x = (\frac{1}{2}) (3 \sqrt{19} - 13), - (\frac{1}{2}) (3 \sqrt{19} + 13)$ $color(brown)(x = (1/2)(3sqrt19 - 13), -(1/2) (3sqrt19 + 13)$

Explanation:

$3 x^{2} - 26 x + 2 = 5 x^{2} + 1$ $3x^2 - 26x+ 2 = 5x^2 + 1$

$5 x^{2} - 3 x^{2} + 26 x = 1$ $5x^2 - 3x^2 + 26x = 1$

$2 (x^{2} + (2 \cdot x \cdot (\frac{13}{2})) = 1$ $2(x^2 + (2 * x * (13/2)) = 1$

$x^{2} + 2 x \cdot (\frac{13}{2}) = \frac{1}{2}$ $x^2 + 2x * (13/2) = 1/2$

Adding ${(\frac{13}{2})}^{2}$ $(13/2)^2$ to both sides,

$x^{2} + 2 x (\frac{13}{2}) + {(\frac{13}{2})}^{2} = \frac{1}{2} + {(\frac{13}{2})}^{2} = (\frac{171}{4})$ $x^2 + 2x (13/2) + (13/2)^2 = 1/2 + (13/2)^2 = (171/4)$

${(x + \frac{13}{2})}^{2} = {(\sqrt{\frac{171}{4}})}^{2}$ $(x + 13/2)^2 = (sqrt(171/4))^2$

$x + \frac{13}{2} = \pm \sqrt{\frac{171}{4}}$ $x + 13/2 = +- sqrt(171/4)$

$x = \pm \sqrt{\frac{171}{4}} - \frac{13}{2}$ $x = +- sqrt(171/4) - 13/2$

$x = (\frac{1}{2}) (3 \sqrt{19} - 13), - (\frac{1}{2}) (3 \sqrt{19} + 13)$ $color(brown)(x = (1/2)(3sqrt19 - 13), -(1/2) (3sqrt19 + 13)$

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How do you solve $3 s^{2} - 26 x + 2 = 5 s^{2} + 1$ $3s^2-26x+2=5s^2+1$ by completing the square?

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How do you solve 3s2−26x+2=5s2+13s^2-26x+2=5s^2+1 by completing the square?

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How do you solve $3 s^{2} - 26 x + 2 = 5 s^{2} + 1$ $3s^2-26x+2=5s^2+1$ by completing the square?