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# Just a quick maths question watch

1. This left me wondering...

Given that I want to complete the square of:

$-x^2-6x+7=0$

Could I just do this to make my life easier?

$-x^2-6x+7=x^2+6x-7$

$x^2+6x-7=0$
2. (Original post by Aklaol)
This left me wondering...

Given that I want to complete the square of:

$-x^2-6x+7=0$

Could I just do this to make my life easier?

$-x^2-6x+7=x^2+6x-7$

$x^2+6x-7=0$
No because

Though you can indeed take out a factor of -1 and complete the square inside the bracket before multiplying through by -1.

So and complete the square inside the brackets
3. (Original post by RDKGames)
No because

Though you can indeed take out a factor of -1 and complete the square inside the bracket before multiplying through by -1.

So and complete the square inside the brackets
Can't I multiply each side by -1 to get rid of the negative coefficient of x?
4. Well yeah but the conventional way is to put brackets around the equation and then put the '-' outside so its -(x^2 + 6x - 7) = 0 and then complete the square so -[(x^2-3x)-7-9] = - [(x^2-3x)-16] = -(x^2 - 3x) + 16
5. (Original post by Aklaol)
Can't I multiply each side by -1 to get rid of the negative coefficient of x?
If you are SOLVING the equation, then you can do it your way. But if you are expressing that quadratic in its completed square form, then you need to take into account the negative. Of course, you can go both ways and see that the completed square form equaling 0 can be written in either form regardless of the -1.
6. (Original post by RDKGames)
If you are SOLVING the equation, then you can do it your way. But if you are expressing that quadratic in its completed square form, then you need to take into account the negative. Of course, you can go both ways and see that the completed square form equaling 0 can be written in either form regardless of the -1.
So in theory, if I was to write this down
$-x^2-6x+7=-(x+3)^2+2$

Would I be wrong?
7. (Original post by Aklaol)
So in theory, if I was to write this down
$-x^2-6x+7=-(x+3)^2+2$

Would I be wrong?
Ah close.

Now if you're solving the quadratic equaling to 0, then the RHS will be equal to 0 - so if you multiply by -1 you will still get the same solutions.
8. (Original post by RDKGames)
Ah close.

Now if you're solving the quadratic equaling to 0, then the RHS will be equal to 0 - so if you multiply by -1 you will still get the same solutions.
And are you positive that I can't do this either?

$-1(-x^2-6x+7)=(0)-1$

(Multiplying each side by -1)

$x^2+6x-7=0$

Completed square:

$(x+3)^2-16$
9. (Original post by Aklaol)
And are you positive that I can't do this either?

$-1(-x^2-6x+7)=(0)-1$

(Multiplying each side by -1)

$x^2+6x-7=0$

Completed square:

$(x+3)^2-16$
Yeah you can. I was just initially pointing out that saying is not correct, and you wouldn't be able to do this if it were just an expression.
10. (Original post by RDKGames)
Yeah you can. I was just initially pointing out that saying is not correct, and you wouldn't be able to do this if it were just an expression.
One last thing, are these statements correct:

$(x+3)^2-16 = x^2+6x-7$

$(x+3)^2-16 \neq -x^2-6x+7$
11. (Original post by Aklaol)
One last thing, are these statements correct:

$(x+3)^2-16 = x^2+6x-7$

$(x+3)^2-16 \neq -x^2-6x+7$
That's right. If you plot them then it would be obvious how they are not the same, but their roots are.
12. (Original post by RDKGames)
That's right. If you plot them then it would be obvious how they are not the same, but their roots are.
Great, thanks.
13. (Original post by RDKGames)
That's right. If you plot them then it would be obvious how they are not the same, but their roots are.
Well actually, if I was to go on to solve this, would I get:

$(x+3)^2-16=0$

$(x+3)^2=16$

$\sqrt{(x+3)^2}=\sqrt{16}$

$x+3=\pm \sqrt{16}$

$x=-3\pm \sqrt{16}$
14. (Original post by Aklaol)
Well actually, if I was to go on to solve this, would I get:

$(x+3)^2-16=0$

$(x+3)^2=16$

$\sqrt{(x+3)^2}=\sqrt{16}$

$x+3=\pm \sqrt{16}$

$x=-2\pm \sqrt{16}$
Should be -3 at the very last line.
15. (Original post by RDKGames)
Should be -3 at the very last line.
LOL I have no idea why I typed -2.
16. It isn't set to zero so you can't just change the signs of one side. You always have to balance it if that makes sense.

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