How do you solve using completing the square method $2 x^{2} - 4 x - 3 = 0$ $2x^2-4x-3=0$ ?

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Answer 1 · 2016-10-08T00:45:22

$2 (x^{2} - 2 x + m - m) = 3$ $2(x^2 - 2x + m - m) = 3$

$m$ $m$ has to make the expression in the parentheses a perfect square. So, $m = 1$ $m = 1$ .

$2 (x^{2} - 2 x + 1 - 1) = 3$ $2(x^2 - 2x + 1 - 1) = 3$

$2 (x^{2} - 2 x + 1) - 2 = 3$ $2(x^2 - 2x + 1) - 2 = 3$

$2 {(x - 1)}^{2} = 5$ $2(x - 1)^2 = 5$

${(x - 1)}^{2} = 2.5$ $(x - 1)^2 = 2.5$

$x - 1 = \pm \sqrt{2.5}$ $x - 1= +-sqrt(2.5$

$x = 1 \pm \sqrt{2.5}$ $x = 1 +-sqrt(2.5)$

Hopefully this helps!

How do you solve using completing the square method 2x2−4x−3=02x^2-4x-3=0?