How do you solve $2 x^{2} - 12 x + 11 = 0$ $2x^2 -12x + 11=0$ by completing the square?

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Answer 1 · 2015-05-13T20:25:16

First divide through by 2 to get:

$x^{2} - 6 x + \frac{11}{2} = 0$ $x^2 - 6x + 11/2 = 0$

Now ${(x - 3)}^{2}$ $(x - 3)^2$ = $x^{2} - 6 x + 9$ $x^2 -6x + 9$

So we can write

$0 = x^{2} - 6 x + \frac{11}{2}$ $0 = x^2 - 6x + 11/2$

$= x^{2} - 6 x + 9 - 9 + \frac{11}{2}$ $= x^2 -6x + 9 - 9 + 11/2$

$= {(x - 3)}^{2} - 9 + \frac{11}{2}$ $= (x-3)^2 - 9 + 11/2$

$= {(x - 3)}^{2} - \frac{18}{2} + \frac{11}{2}$ $= (x - 3)^2 - 18/2 + 11/2$

$= {(x - 3)}^{2} - \frac{7}{2}$ $= (x - 3)^2 - 7/2$

Adding $\frac{7}{2}$ $7/2$ to both sides we get

${(x - 3)}^{2} = \frac{7}{2}$ $(x - 3)^2 = 7/2$

So $(x - 3) = \pm \sqrt{\frac{7}{2}}$ $(x - 3) = +- sqrt(7/2)$

Add 3 to both sides to get

$x = 3 \pm \sqrt{\frac{7}{2}} = 3 \pm \frac{\sqrt{7}}{\sqrt{2}}$ $x = 3 +- sqrt(7/2) = 3 +- sqrt(7)/sqrt(2)$

If you prefer, $\sqrt{\frac{7}{2}} = \sqrt{\frac{14}{4}} = \frac{\sqrt{14}}{2}$ $sqrt(7/2) = sqrt(14/4) = sqrt(14)/2$ , so

$x = 3 \pm \frac{\sqrt{14}}{2}$ $x = 3 +- sqrt(14)/2$

In general,

$a x^{2} + b x + c = a {(x + \frac{b}{2 a})}^{2} + (c - \frac{b^{2}}{4 a})$ $ax^2 + bx + c = a(x + b/(2a))^2 + (c - b^2/(4a))$

Hence $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ has solutions

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +-sqrt(b^2 - 4ac))/(2a)$

How do you solve 2x2−12x+11=02x^2 -12x + 11=0 by completing the square?