Using the complete the sqaure formula.. $2x^2 - 4x +1 = 0$

Question

So Im using the complete square method and i was just wondering where am i going wrong. I'm solving this $2x^2 - 4x +1 = 0$

So i am using this rules. $$ax^2 + bx +c = 0$$

1. subract c from both sides; $ax^2 + bx = -c$

2. divide by a $ x^2 + \frac{b}{a}x = \frac{-c}{a}$

3. add $(\frac{1}{2} \cdot co-effiecient of x)^2$ on both sides;

so; $2x^2 - 4x = -1$

going to step 2

$\frac{2}{2}x^2 - \frac{4}{2}^2 = \frac{-1}{2} + \frac{2}{2}^2$

this will be:

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

this will be:

$x - 1 = \sqrt{\frac{-1}{2} + 1}$

after $l.c.m$

= $x = \pm\frac{1}{2} $

Where am i going wrong?

Answer 1

Let's get you started: $$ 2x^2 - 4x + 1 = 0\\ 2x^2 - 4x + 2 = 1\\ 2(x-1)^2 = 1 $$ Can you take it from there?

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In response to your edits:

You found that $$ (x-1)^2 = -\frac 12 + 1 $$ Or in other words, $$ (x-1)^2 = \frac 12 $$ From there, we have $$ x - 1 = \pm \sqrt{\frac 12} = \pm \frac{\sqrt 2}{2} $$ Adding $1$ to both sides, we have $$ x = 1 \pm \frac{\sqrt 2}2 $$

Answer 2

$$2x^2-4x+1=0$$ $$2(x^2-2x)+1=0$$ $$2(x^2-2x+1)+1=2*1$$ $$2(x-1)^2=1$$ $$(x-1)^2=\frac{1}{2}$$ $$x-1=\pm \sqrt{\frac{1}{2}}$$ $$x=1\pm \sqrt{\frac{1}{2}}$$ $$x=1\pm \frac{\sqrt{2}}{2}$$

Answer 3

Your answer is not correct. You can see why by inputting your $x$-values into your original polynomial $$f(x)=2x^2-4x+1.$$ If $f(x)=0$, as in your problem, and you input your calculated values of $x_1=4$ and $x_2=-4$ you obtain $f$ values of $17$ and $49$, respectively. Because you want $f(x)=0$, these values are not correct.

To use the complete-the-square method, you must first set $f(x)$ equal to zero, like you did.

$$2x^2-4x+1=0. $$ I like to simplify the equation so that the first term has a coefficient of $1$ in front. Thus, $$x^2-2x+\frac{1}{2}=0. $$ Isolate the terms with variables on one side of the equation and put any constants on the other. $$x^2-2x=-\frac{1}{2}. $$ Now, we need to add a coefficient to both sides of the equation that will make the left side become a perfect square. Because $$(x-1)(x-1)=x^2-2x+1, $$ we add $1$ to both sides. Thus, $$x^2-2x+1=\frac{1}{2},$$ or, identically, $$(x-1)^2=\frac{1}{2}. $$ We solve for $x$: $$x-1=\pm\sqrt{\frac{1}{2}}=\pm\frac{\sqrt{2}}{2}.$$ Consequently, $$x=1\pm \frac{\sqrt{2}}{2}.$$

Answer 4

From your equation we have, simply adding and subtracting $1$ to the equation to get: $$\begin{align}2x^2 - 4x +1 &= 0 \\ 2x^2 -4x + 2 - 1 &= 0\end{align}$$

But you know that $2(x^2 - 2x + 1) = 2(x-1)^2$ so your equation turns into $$2(x-1)^2 - 1 = 0$$ Now that you have only on instance of $x$ you can solve for it. Add $1$ and divide by $2$ on both sides to get $$(x-1)^2 = \frac{1}{2}$$ - now take the square root, but don't forget to add $\pm$ to get $$x-1 = \pm \frac{1}{\sqrt{2}}$$ You end up with $$\bbox[10px, border:solid blue 1px]{x = 1 \pm \frac{1}{\sqrt{2}}}$$