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

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

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Answer 1 · 2015-04-05T10:04:42

We know that ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2 = x^2+2ax+a^2$ . Since your equation starts with $x^{2} - 4 x$ $x^2-4x$ , it means that we must find a value of $a$ $a$ such that $2 a x = - 4 x$ $2ax=-4x$ , which means $a = - 2$ $a=-2$ .

If $a = - 2$ $a=-2$ , then we have ${(x - 2)}^{2} = x^{2} - 4 x + 4$ $(x-2)^2 = x^2-4x+4$ . You can see that your equation $x^{2} - 4 x + 2$ $x^2-4x+2$ is very close: we only need to add $2$ $2$ .

So, adding $2$ $2$ to both sides, we get $x^{2} - 4 x + 4 = 2$ $x^2-4x+4=2$ , which means

${(x - 2)}^{2} = 2$ $(x-2)^2 = 2$ . Extracting square roots at both sides, we get $x - 2 = \pm \sqrt{2}$ $x-2= \pm\sqrt(2)$ , which leads us to the solutions

$x_{1, 2} = 2 \pm \sqrt{2}$ $x_{1,2} = 2 \pm \sqrt(2)$

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

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