Just a quick maths question

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

-x^2-6x+7\not= x^2+6x-7

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

So

-x^2-6x+7=-(x^2+6x-7)

and complete the square inside the brackets

Reply 2

Original post by RDKGames

No because

-x^2-6x+7\not= x^2+6x-7

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

So

-x^2-6x+7=-(x^2+6x-7)

and complete the square inside the brackets

Can't I multiply each side by -1 to get rid of the negative coefficient of x?

Reply 3

azizadil1998

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

Reply 4

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.

Reply 5

Original post by RDKGames

So in theory, if I was to write this down
$-x^2-6x+7=-(x+3)^2+2$

Would I be wrong?

Reply 6

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.

-x^2-6x+7=-(x^2+6x-7)=-[(x+3)^2-9-7]=16-(x+3)^2

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.

Reply 7

Original post by RDKGames

Ah close.

-x^2-6x+7=-(x^2+6x-7)=-[(x+3)^2-9-7]=16-(x+3)^2

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:

$(x+3)^2-16$

Reply 8

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

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

is not correct, and you wouldn't be able to do this if it were just an expression.

Reply 9

Original post by RDKGames

Yeah you can. I was just initially pointing out that saying

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

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:

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

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

Reply 10

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.

Reply 11

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.

Reply 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.

Well actually, if I was to go on to solve this, would I get:

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

$(x+3)^2=16$

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

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

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

(edited 7 years ago)

Reply 13