How do you solve using the completing the square method $10 x^{2} = 4 x + 7$ $10x^2 = 4x + 7$ ?

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How do you solve using the completing the square method $10 x^{2} = 4 x + 7$ $10x^2 = 4x + 7$ ?

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Answer 1 · 2017-07-09T20:22:19

$x_{1} = \frac{2 + \sqrt{74}}{10} and x_{2} = \frac{2 - \sqrt{74}}{10}$ $x_1=(2+sqrt(74))/10 and x_2=(2-sqrt(74))/10$

Explanation:

$10 x^{2} = 4 x + 7$ $10x^2=4x+7$

$10 x^{2} - 4 x - 7 = 0$ $10x^2-4x-7=0$

$100 x^{2} - 40 x - 70 = 0$ $100x^2-40x-70=0$

$100 x^{2} - 40 x + 4 - 74 = 0$ $100x^2-40x+4-74=0$

${(10 x - 2)}^{2} - {(\sqrt{74})}^{2} = 0$ $(10x-2)^2-(sqrt(74))^2=0$

${(10 x - 2)}^{2} = {(\sqrt{74})}^{2}$ $(10x-2)^2=(sqrt(74))^2$

Hence $x_{1} = \frac{2 + \sqrt{74}}{10}$ $x_1=(2+sqrt(74))/10$ and $x_{2} = \frac{2 - \sqrt{74}}{10}$ $x_2=(2-sqrt(74))/10$

Answer 2 · 2017-07-09T20:56:05

$x = \frac{1}{5} \pm \sqrt{\frac{37}{50}}$ $x = 1/5 +- sqrt(37/50)$

Explanation:

$10 x^{2} - 4 x = 7$ $10x^2 - 4x = 7$
Divide both sides by 10:
$x^{2} - \frac{4 x}{10} = \frac{7}{10}$ $x^2 - (4x)/10 = 7/10$
$x^{2} - (2 \frac{x}{5}) = \frac{7}{10}$ $x^2 - (2x/5) = 7/10$
$(x^{2} - \frac{2 x}{5}) + \frac{1}{25} = \frac{7}{10} + \frac{1}{25}$ $(x^2 - (2x)/5) + 1/25 = 7/10 + 1/25$
${(x - \frac{1}{5})}^{2} = \frac{37}{50}$ $(x - 1/5)^2 = 37/50$
$(x - \frac{1}{5}) = \pm \sqrt{\frac{37}{50}}$ $(x - 1/5) = +- sqrt(37/50)$
$x = \frac{1}{5} \pm \sqrt{\frac{37}{50}}$ $x = 1/5 +- sqrt(37/50)$

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How do you solve using the completing the square method $10 x^{2} = 4 x + 7$ $10x^2 = 4x + 7$ ?

2 Answers

Explanation:

Explanation:

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How do you solve using the completing the square method $10 x^{2} = 4 x + 7$ $10x^2 = 4x + 7$ ?