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

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Answer 1 · 2015-05-13T20:25:16

First divide through by 2 to get:

$x^2 - 6x + 11/2 = 0$

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

So we can write

$0 = x^2 - 6x + 11/2$

$= x^2 -6x + 9 - 9 + 11/2$

$= (x-3)^2 - 9 + 11/2$

$= (x - 3)^2 - 18/2 + 11/2$

$= (x - 3)^2 - 7/2$

Adding $7/2$ to both sides we get

$(x - 3)^2 = 7/2$

So $(x - 3) = +- sqrt(7/2)$

Add 3 to both sides to get

$x = 3 +- sqrt(7/2) = 3 +- sqrt(7)/sqrt(2)$

If you prefer, $sqrt(7/2) = sqrt(14/4) = sqrt(14)/2$ , so

$x = 3 +- sqrt(14)/2$

In general,

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

Hence $ax^2 + bx + c = 0$ has solutions

$x = (-b +-sqrt(b^2 - 4ac))/(2a)$

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