How do you solve the quadratic equation by completing the square: $x^{2} - 4 x + 2 = 0$ $x^2-4x+2=0$ ?

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Answer 1 · 2015-07-19T10:40:42

Complete the square to find ${(x - 2)}^{2} = x^{2} - 4 x + 4 = 2$ $(x-2)^2 = x^2-4x+4 = 2$

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

Add $2$ $2$ to both sides to get:

$2 = x^{2} - 4 x + 4 = {(x - 2)}^{2}$ $2 = x^2-4x+4 = (x-2)^2$

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

Add $2$ $2$ to both sides to get:

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

In the general case:

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

from which we can derive the quadratic formula for solutions of $a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$ :

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

Notice the $\frac{b}{2 a}$ $b/(2a)$ term that gives us:

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

How do you solve the quadratic equation by completing the square: x2−4x+2=0x^2-4x+2=0?