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I have been trying to learn some pre-calculus stuff in advance for next year (attempting a university paper)

I am trying to solve a completing the square equation but can't see where I am going wrong :

I go through the following process:

$$2x^{ 2 } - 8x+1=0$$

Subtract 1 from both sides:

$$2x^{ 2 } - 8x\quad =-1$$

add $(-8/2)^2$ to both sides

$$2x^{ 2 } - 8x + (\frac {-8}{2})^2 = -1 + (\frac { -8 }{ 2 })^2$$

which gives me $$2x^{ 2 } - 8x + (-4)^2 = -1 + (-4)^2 = -1 + 16 = 15$$

Now I divide both sides by 2 and get

$$x^{ 2 } - 4x - 2^2 = \frac { 15 }{ 2 } $$

Now I have

$$(x - 2)^{ 2 } = \frac { 15 }{ 2 }$$

But if I expand $$(x - 2)^{ 2 } $$

It is $$(x^2 -4x + 4)$$ So at this point I see I have gone wrong somewhere..

Updated…..

turns out my answer should be $$x={ 2 }\pm\sqrt { \frac { 7 }{ 2 } } $$

however I can't see how they got to this.

the working shows $$ x^2 -4x +2^2 = -1/2 + 2^2$$

I am unsure how they for a positive $2^2$ in the left hand side

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  • $\begingroup$ $(-4)^2=4^2$... You added the number and showed it as positive on the RHS, but then forgot to square away the negative on the LHS. $\endgroup$ – abiessu Nov 2 '14 at 7:55
  • $\begingroup$ You need to divide $-8x$ by $2\sqrt{2x^2}$, and then square it to find the last term $$2x^{ 2 } - 8x + (\frac {-8}{2})^2 = -1 + (\frac { -8 }{ 2 })^2$$ should read $$2x^{ 2 } - 8x + (\frac {-8}{2\sqrt{2}})^2 = -1 + (\frac { -8 }{ 2\sqrt{2} })^2$$ $\endgroup$ – John Joy Nov 2 '14 at 16:18
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Note that adding $(-4)^2 = 16$ to $2x^2 - 8x$ does not form a perfect square.

Also, when you divided the expression $2x^2 - 8x + (-4)^2 = 2x^2 - 8x + 16$ by $2$, you should have obtained $x^2 - 4x + 8$ rather than $x^2 - 4x - 4$. You divided $(-4)^2$ by $-4$ when you obtained $x^2 - 4x - 2^2 = x^2 - 4x - 4$.

To complete the square on

$$2x^2 - 8x = -1$$

we divide each side of the equation by $2$ to obtain

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

We want to form a perfect square on the left hand side of the equation. We can do so by obtaining an expression of the form

$$x^2 + 2kx + k^2 = (x + k)^2$$

In the expression $x^2 - 4x$, $2k = -4$, so $k = -2$. Thus, we must add $k^2 = 4$ to each side of the equation $$x^2 - 4x = -\frac{1}{2}$$ to obtain \begin{align*} x^2 - 4x + 4 & = -\frac{1}{2} + 4\\ (x - 2)^2 & = \frac{7}{2}\\ x - 2 & = \pm \sqrt{\frac{7}{2}}\\ x - 2 & = \pm \frac{\sqrt{7}}{\sqrt{2}}\\ x - 2 & = \pm \frac{\sqrt{7}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\ x - 2 & = \pm \frac{\sqrt{14}}{2}\\ x & = 2 \pm \frac{\sqrt{14}}{2}\\ x & = \frac{4 \pm \sqrt{14}}{2} \end{align*}

Another option is to transform the equation $$2x^2 - 8x = -1$$ by multiplying both sides of the equation by $2$ to obtain the equation $$4x^2 - 16x = -2$$ We want to obtain a perfect square on the left hand side of the equation, which we can do if we obtain an expression of the form $$(2x + k)^2 = 4x^2 + 4kx + k^2$$ In the expression $4x^2 - 16x$, $4k = -16$, so $k = -4$. Thus, we must add $k^2 = 16$ to each side of the equation $$4x^2 - 16x = -2$$ which yields \begin{align*} 4x^2 - 16x + 16 & = -2 + 16\\ (2x - 4)^2 & = 14\\ 2x - 4 & = \pm\sqrt{14}\\ 2x & = 4 \pm \sqrt{14}\\ x & = \frac{4 \pm \sqrt{14}}{2} \end{align*} which is equivalent to the previous result.

The idea behind completing the square was introduced by Muhammad ibn-Musa al-Khowarizmi in his text Hisab al-jabr wa'l Muqubalah (the first general text on algebra). He wanted to solve the problem $x^2 + 10x = 39$, so he viewed $x^2$ as the area of a square of side length $x$ and $10x$ as the area of two $5$ by $x$ rectangles, which he attached to the sides of the square, as shown below. He completed the square by adding the area of the $5$ by $5$ rectangle shown with dashed sides.

completing_the_square

By adding $25$ to each side, he obtained the area of a square with side length $x + 5$.

\begin{align*} x^2 + 10x & = 39\\ x^2 + 10x + 25 & = 39 + 25\\ (x + 5)^2 & = 64\\ x + 5 & = \pm 8\\ x & = 3, -13 \end{align*}

Since he was interested in the geometric solution, he discarded the algebraically valid solution $x = -13$. However, the diagram should give you a sense of why we add $\left(\frac{b}{2}\right)^2 = \frac{b^2}{4}$ to the expression $x^2 + bx$ to complete the square.

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  • $\begingroup$ I can follow up to the part where you get x -2 = plus minus sort(7/2). But after that I am unsure why you multiply it by sort (2/2). And then I'm also unsure of what law or rule applies when multiplying sqrt 7 * sqrt 2 ? I realise they both can be viewed as begin to the power of 1/2 but I'm unsure how they multiply due to having different bases (i.e 7 and 2) $\endgroup$ – user3754366 Nov 2 '14 at 20:19
  • $\begingroup$ What I am doing is called rationalizing the denominator. If $a$ and $b$ are positive real numbers, then $$\sqrt{ab} = \sqrt{a}\sqrt{b}$$ and $$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$ In particular, $\sqrt{a}\sqrt{a} = a$. Thus, $$\sqrt{\frac{7}{2}} \cdot \sqrt{\frac{2}{2}} = \frac{\sqrt{7}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{14}}{2}$$ Now the denominator is rational. Rationalizing the denominator used to be important when people had to consult tables to approximate square roots. Dividing by $2$ is easier than dividing by $\sqrt{2}$. $\endgroup$ – N. F. Taussig Nov 2 '14 at 20:32
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It seems you have applied the "add the square of half the linear coefficient" idea. This only works if you have coefficient $1$ on the $x^2,$ so start by factoring the $2$ out in your example, getting to $2(x^2-4x+1/2).$ The method should now be applied to $x^2-4x+1/2$ and then remember to insert the extra factor of $2.$

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  • $\begingroup$ ahhh …that's good to know. I didn't realise it only worked with a coefficient of 1 on the x^2. thanks $\endgroup$ – user3754366 Nov 2 '14 at 8:02
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Actually, your mistake is when you divide both sides by two. What you wrote means that $\frac{-4^2}{2}=-2^2$ but this is not true since $-4^2=16$ and $16/2=8\neq-2^2$

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  • $\begingroup$ I've edited it as I believe it should have read $$\frac {(-4/2)^2}$$ $\endgroup$ – user3754366 Nov 2 '14 at 8:17

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