How do you solve $2 x^{2} - 12 x + 7 = 5$ $2x^2 - 12x + 7 = 5$ ?

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How do you solve $2 x^{2} - 12 x + 7 = 5$ $2x^2 - 12x + 7 = 5$ ?

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Answer 1 · 2018-03-28T23:15:40

$x = 3 \pm 2 \sqrt{2}$ $x = 3+-2sqrt(2)$

Explanation:

Given:

$2 x^{2} - 12 x + 7 = 5$ $2x^2-12x+7=5$

Subtract $5$ $5$ from both sides to get:

$2 x^{2} - 12 x + 2 = 0$ $2x^2-12x+2=0$

Divide both sides by $2$ $2$ to get:

$0 = x^{2} - 6 x + 1$ $0 = x^2-6x+1$

$0 = x^{2} - 6 x + 9 - 8$ $color(white)(0) = x^2-6x+9-8$

$0 = {(x - 3)}^{2} - {(2 \sqrt{2})}^{2}$ $color(white)(0) = (x-3)^2-(2sqrt(2))^2$

$0 = ((x - 3) - 2 \sqrt{2}) ((x - 3) + 2 \sqrt{2})$ $color(white)(0) = ((x-3)-2sqrt(2))((x-3)+2sqrt(2))$

$0 = (x - 3 - 2 \sqrt{2}) (x - 3 + 2 \sqrt{2})$ $color(white)(0) = (x-3-2sqrt(2))(x-3+2sqrt(2))$

Hence:

$x = 3 \pm 2 \sqrt{2}$ $x = 3+-2sqrt(2)$

Answer 2 · 2018-03-29T00:34:43

$x = 3 \pm 2 \sqrt{2}$ $x = 3 +- 2sqrt2$

Explanation:

2x^2 - 12x + 2 = 0
x^2 - 6x + 1 = 0
Use the improved quadratic formula (Socratic search):
$D = d^{2} = b^{2} - 4 a c = 36 - 4 = 32 = 2 (16)$ $D = d^2 = b^2 - 4ac = 36 - 4 = 32 = 2(16)$
$d = \pm 4 \sqrt{2}$ $d = +- 4sqrt2$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - (- \frac{6}{2}) \pm \frac{4 \sqrt{2}}{2}$ $x = -b/(2a) +- d/(2a) = - (-6/2) +- (4sqrt2)/2$
$x = (3 \pm 2 \sqrt{2})$ $x = (3 +- 2sqrt2)$

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How do you solve $2 x^{2} - 12 x + 7 = 5$ $2x^2 - 12x + 7 = 5$ ?

2 Answers

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How do you solve $2 x^{2} - 12 x + 7 = 5$ $2x^2 - 12x + 7 = 5$ ?