How do you solve by completing the square: $2 x^{2} + 4 x + 1 = 0$ $2x^2 + 4x +1 = 0$ ?

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How do you solve by completing the square: $2 x^{2} + 4 x + 1 = 0$ $2x^2 + 4x +1 = 0$ ?

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Answer 1 · 2018-04-30T03:08:16

Add one and then divide by 2

Explanation:

First add 1 to each side
$2 x^{2} + 4 x + 2 = 1$ $2x^2+4x+2=1$
Then divide each side by two:
$x^{2} + 2 x + 1 = \frac{1}{2}$ $x^2+2x+1=1/2$
Then you see that $x^{2} + 2 x + 1 = {(x + 1)}^{2}$ $x^2+2x+1=(x+1)^2$
${(x + 1)}^{2} = \frac{1}{2}$ $(x+1)^2=1/2$
Take the square root of each side:
$x + 1 = \frac{1}{\sqrt{2}}$ $x+1=1/sqrt(2)$
$x = \frac{1}{\sqrt{2}} - 1 = \frac{\sqrt{2} - 2}{2}$ $x=1/sqrt(2)-1=(sqrt(2)-2)/2$

Answer 2 · 2018-04-30T03:15:20

$x = - 1 - \frac{\sqrt{2}}{2}$ $x=-1-sqrt(2)/2$ , and $x = - 1 + \frac{\sqrt{2}}{2}$ $x=-1+sqrt(2)/2$

Explanation:

First divide both sides of the equation by 2.

Equation 1
$x^{2} + 2 x + \frac{1}{2} = 0$ $x^2+2x+1/2=0$

Now add $\frac{1}{2}$ $1/2$ to both sides of the equation. (see footnote)

$x^{2} + 2 x + 1 = \frac{1}{2}$ $x^2+2x+1=1/2$

Now we can factor the left-hand side of the equation.

${(x + 1)}^{2} = \frac{1}{2}$ $(x+1)^2=1/2$

Now let's write the equation as the difference of two squares.

${(x + 1)}^{2} - \frac{1}{2} = 0$ $(x+1)^2-1/2=0$

The square root of $\frac{1}{2}$ $1/2$ is $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$ , so we can factor the above expression as

$(x + 1 + \frac{\sqrt{2}}{2}) (x + 1 - \frac{\sqrt{2}}{2}) = 0$ $(x+1+sqrt(2)/2)(x+1-sqrt(2)/2)=0$

Therefore the solution to this equation is

$x = - 1 - \frac{\sqrt{2}}{2}$ $x=-1-sqrt(2)/2$ , and $x = - 1 + \frac{\sqrt{2}}{2}$ $x=-1+sqrt(2)/2$

FOOTNOTE:

Why did I pick $\frac{1}{2}$ $1/2$ to add to both sides of Equation 1? I looked at the coefficient for the $x$ $x$ (linear) term in Equation 1 which is 2. I divided 2 by 2 and then squared the result. ${(\frac{2}{2})}^{2} = 1$ $(2/2)^2=1$ is what I wanted the THIRD term on the left hand side of Equation 1 to be. This is why I added $\frac{1}{2}$ $1/2$ to both sides. If, for example, the coefficient in for the $x$ $x$ term had been 5 instead of 2, then I would have wanted the third term on the left-hand side of the equation to be ${(\frac{5}{2})}^{2} = \frac{25}{4}$ $(5/2)^2=25/4$ and I would have needed to add $\frac{23}{4}$ $23/4$ to both sides of the equation.

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How do you solve by completing the square: $2 x^{2} + 4 x + 1 = 0$ $2x^2 + 4x +1 = 0$ ?

2 Answers

Explanation:

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