# Complete the square for $f(x) = 2x^2 + 4x - 6$

I'm studying for a math test. This is the question:

$f(x) = 2x^2 + 4x - 6$. complete the square.

This is how much I get out of the question: $$2x^2 + 4x - 6$$ $$2(x^2 + 2x - 3)$$ $$2(x^2 + 2x + 1^2 - 1^2 - 3)$$ $$2((x + 1)^2 - 4)$$

But I get stuck here. Can someone complete it and explain?

I'm sorry i forgot a important piece off the question it says: Complete the square and derived to get the minimum value off $f(x)$

You are more-or-less done. I don't know how you are asked to leave your answer, but there are three possibilities.

1. You could leave it as it is: $2x^2 + 4x-6 \equiv 2[(x+1)^2-4]$.
2. You could expand the factor of two: $2x^2 + 4x - 6 \equiv 2(x+1)^2 - 8$.
3. You could take the two inside the square: $2x^2 + 4x -6 \equiv (\sqrt{2}x+\sqrt{2})^2-8.$

EDIT:

You added the fact that you then must find the minimum value of the function. Since $\operatorname{f}(x) \equiv 2(x+1)^2 - 8$ you find that $\operatorname{f}(x) \ge -8$ for all $x$. This is because $2(x+1)^2 \ge 0$ for all $x$. The function takes its minimum value when $2(x+1)^2 = 0$, i.e. when $x=-1$.

• Indeed, it would be a very harsh marker if you (@Ernes Filipovic) were to not gain full marks for simply leaving it as it is. (Also, $3$ is odd, and would only really make sense (for some value of "make sense") if $2$ was a square number (so, for example, you had $9((x+1)^2+4)$ instead).) – user1729 May 17 '13 at 15:17

Given:

$f(x) = 2x^2 + 4x - 6$. complete the square.

$f(x)=2x^2+4x-6$

$f(x)=2x^2+4x-6$

$f(x)=2(x^2+2x)-6$

$f(x)=2[x^2+\dfrac{2x}{2}+(1)^2]-6$

$f(x)=[2(x+1)^2 -1]-6$

$f(x)=2(x+1)^2-8$ which is the same as: $f(x) = 2x^2 + 4x - 6$

This can be confirmed by using a graphing calculator such as: https://www.desmos.com/

$f(x) = 2x^2 + 4x - 6$. complete the square.

This is how much I get out of the question: $$2x^2 + 4x - 6$$ $$2(x^2 + 2x - 3)$$ For ease of understanding at this stage ignore the 2 and consider

$$(x^2 + 2x - 3)$$ Next step $$(x+1)^2=x^2+2x+1$$ Here we have divided the coefficient of the x term by $\frac{1}{2}$ because when we times out we get the term twice as you have just seen there.

Next note we now have 1 instead of -3 thus to balance this we -4. That is, $$(x+1)^2+1-4 = (x+1)^2-3$$

Now a good point about completing the square is we can find the minimun point and value easily. Note that with the right hand side RHS we have a squared term which contains the $x$ since all squared numbers are greater than or equal to zero the smallest we can make this term is $0$ and this is done at $x=-1$.

Now we have this point we can substitute $x=-1$ into our earlier equation:

$$2(x^2 + 2x - 3)$$

to get

$$2((-1)^2 + 2(-1) - 3) = -8$$

Rewrite your expression as $$2x^2+4x+b-6-b$$ for any $b>0$. To close the square you need the expression in the form $$(\alpha x +\beta)^2-(6+b)=\alpha^2x^2+2 \alpha \beta x+\beta^2-(6+b)$$ You get immediately $\alpha=|\sqrt{2}|,\beta=|\sqrt{b}|$, hence your expression is $$(|\sqrt{2}|x+|\sqrt{b}|)^2-(6+b)$$ If you want you can set $b=2$ to get $$2(x+1)^2-(6+2)$$

• Irrational constant term? – Erick Wong May 17 '13 at 16:43
• This should work for any $b \in \mathbb{R}^{+}$ – Alex May 17 '13 at 17:25
• What I mean is how did the constant term become irrational, when it used to be $-6$? – Erick Wong May 17 '13 at 17:28
• I rewrote $-6=-6+b-b$ – Alex May 17 '13 at 17:31
• Think about it. How can $2(x+1)^2 - (6+\sqrt{2})$ possibly be equal to $2x^2 + 4x - 6$? There is an obvious typo. – Erick Wong May 17 '13 at 19:42

Well, I'll first give an example to motivate the method. Think for example in $(x+1)^2$ if you expand this normally you'll get $x^2+2x+1$, notice that this is square of the first term plus twice the first term times the second term plus the second term squared. So if we have an arbitrary expression like your's it'll be possible to extract from it one expression that is the square of the sum of two numbers plus another number.

You have $f(x) = 2x^2+4x-6$, so divide this by $2$ to get $f(x)/2 = x^2+2x-3$, the reason I did this will soon appear. Now look just on the RHS, if we want to write this as $(x+\text{something})^2$ we know that it'll have before anything else the first term squared, that is, $x^2$, but look there! You already have $x^2$ in your expression (and it's exactly the first term squared because we divided by $2$ so that only $x^2$ remains).

Now, you have $2x$, remember tha the second term appearing after expanding the square should be twice the first term times the second term, now you have there the second term $2x$, and you know this is the second term because it has the first term times something. Note that in the expansion, only in this place appears the first term times something without being squared. Now, since it is twice the first term times the second term, and sice $x$ is the first term, if we divide this by $2\cdot x$ (twice the first term) we'll get the second term! Now, doing this we have that the second term must by $1$.

Now, we have there a minus $3$, not the square of the second term as expected. Now we fix that completing the square, we simply sum and subtract the second term squared (and it won't alter anything since summing and subtracting together aims to summing zero), so we write:

$$f(x)/2 = x^2+2x +1 - 1 - 3$$

Now, the thing $(x^2+2x+1)$ will be just $(x+1)^2$ and $-1-3 = -4$ so that we have in the end $f(x)/2 = (x+1)^2 - 4$ and so $f(x) = 2(x+1)^2 - 8$. You can easily check that's the same thing you had before:

$$f(x) = 2(x^2+2x+1)-8 = 2x^2+4x+2 - 8 = 2x^2 + 4x -6$$

I hope it helps you out. My suggestion is that you try to repeat this argument to another problem so that you get better practice.