Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$?

Question

I'm confused about a question I posted this morning.

I am trying to understand if $-6(x-\frac{1}{2})^2$ can be rewritten as $-\frac{3}{2}(2x-1)^2$?

I tried multiplying out the expression $-6(x-\frac{1}{2})^2$ to a polynomial form $36x^2-36x+9$ but that didn't take me closer to understanding my goal.

I noticed that I can remove the fraction inside $-6(x-\frac{1}{2})^2$ by doubling the contents:

$-6(x-\frac{1}{2})^2$ <> $-6(2x-1)^2$ # used <> for does not equal

I don't think I can simply half the factor -6 to get $-3(2x-1)^2$

As you can no doubt see, I am confused.

Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$ ?

If it does could someone show me how to transform from $-6(x-\frac{1}{2})^2$ to $-\frac{3}{2}(2x-1)^2$ in granular baby steps?

score 5 · Answer 1 · answered Jul 28 '20 at 20:59

5

We have

$$-6\left(x-\frac{1}{2}\right)^2=-\frac 6{\color{blue}4}\cdot \color{blue}4\cdot\left(x-\frac{1}{2}\right)^2=-\frac 3{\color{blue}2}\cdot \color{blue}{2^2}\cdot\left(x-\frac{1}{2}\right)^2=\\=-\frac 32\left(\color{blue}2\cdot x-\color{blue}2\cdot\frac{1}{2}\right)^2=-\frac 32(2x-1)^2$$

answered Jul 28 '20 at 20:59

user

154,566

1

Thank you for the answer, I know this is basic stuff to most on here. I switched to accepting Lee's answer, I was just able to follow along a bit easier. Thank you for taking the time to answer all the same. – Doug Fir Jul 28 '20 at 21:37
@DougFir You are welcome! That's nice now it is clear. Bye – user Jul 28 '20 at 21:40

score 4 · Accepted Answer · 2020-07-28T21:35:54.450

4

Look at my answer, it's quite simple to prove $$-6\left(x-\frac{1}{2}\right)^2=-6\left(\frac{2x-1}{2}\right)^2=-6\frac{\left(2x-1\right)^2}{2^2}=-6\frac{\left(2x-1\right)^2}{4}=-\frac 32(2x-1)^2$$

edited Jul 28 '20 at 21:35

answered Jul 28 '20 at 21:30

I don't fully follow progression from step 2 to 3? How did you get a new denominator of 4? – Doug Fir Jul 28 '20 at 21:33
1

it's just square of $2$ i.e. $2^2=4$ is a new denominator for step 3 – Jul 28 '20 at 21:33
1

This answer did just read much simpler to me. Thank you. – Doug Fir Jul 28 '20 at 21:37
@DougFir: I have added one more step . it is now clearer to you – Jul 28 '20 at 21:37

score 2 · Answer 3 · answered Jul 28 '20 at 21:07

2

$$-6\left(x-\frac{1}{2} \right)^2 = -\frac{3}{2}\left(2x-1 \right)^2$$ $$-6\left( x^2 -x +\frac{1}{4}\right) = -\frac{3}{2} \left(4x^2-4x+1 \right)$$ $$-6x^2+6x - \frac{6}{4} = -6x^2+6x -\frac{3}{2} $$

answered Jul 28 '20 at 21:07

zkutch

13,410

score 1 · Answer 4 · answered Jul 28 '20 at 21:00

Your confusion stems from the fact that once you factor out the $\frac{1}{2}$ (aka double the inside), you still need to square it in order to take it outside the parenthesis. So you are correct in what you are trying to do, but then $(\frac{1}{2})^2 = \frac{1}{4}$ which, when multiplied by $-6$, does give you $-\frac{3}{2}$.

Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$?

4 Answers4