Can you explain this to me (involves real roots and completing the square)?

The question is:
"Given that the equation x^2 + 4x + c =0 has unequal real roots, find the range of possible values of c"

From earlier on in the question, I know that x^2 + 4x + c = (x+2)^2 -4 + c =0

I get the (x+2)^2 -4 + c =0 because the questions earlier stages involved completing the square.

However, in the solution, (x+2)^2 -4 + c =0 then goes to (x+2)^2 = 4-c

But it then goes to 4-c > 0.

What happened to the (x+2)^2? Where did that go? And why is it >0?

Thanks

Dunno dude, in any case the extra bit on the end when you complete the square is the turning point of that parabola, that's as much as i know.

To answer these question i always use the discriminant.

$b^2 -4ac$

unequal roots supposedly means discriminant>0

so $4^2 -4\times 1 \times c > 0$

$16-4c>0$

$4-c>0$

Thanks, that clears it up. I would still like to know in their method, how they suddenly turned the (x+2)^2 = 4-c into 4-c>0

Thanks, that clears it up. I would still like to know in their method, how they suddenly turned the (x+2)^2 = 4-c into 4-c>0

In their method, they seem to start completing the square (for some reason, I cannot work out why) but suddenly skip to the bit where b^2-4ac>0 (for unequal roots).

I will just ignore the mark scheme on this one and use your method. Your method of doing it is so much easier, as you already have the equation given in the question of x^2 + 4x +c =0, so you just apply the rule of unequal roots with the discriminant being b^2-4ac>0, so you then solve it to find c.

Am I right in thinking this ^ about how your method works and is correct?? Can you just clarify if i am right if I were doing that for this question?

$\left(x+2\right)^2=4-c \iff x+2=\sqrt{4-c}$ so if $4-c=0$ then there are equal roots and if $4-c<0$ then there are no real roots, hence for unequal real roots $4-c>0$,

Yeah, but how in the mark scheme, how did they suddenly go from (x+2)^2 = 4-c

to

4-c>0?

I don't get it.

I just showed you, if you square root both sides then the right hand side must be greater than zero to satisfy the question.

$\left(x+2\right)^2=4-c \iff x+2=\sqrt{4-c}$ so if $4-c=0$ then there are equal roots and if $4-c<0$ then there are no real roots, hence for unequal real roots $4-c>0$,

When you complete the square the extra bit is the y-co-ordinate. So when $(x+2)^2=0$ the value of x will be -2 so when we have our x co-ordinate we need the y which is the 4-c, the extra bit on the end. Then we can use the "unequal roots" and set it >0

Still, using the discriminant tells us whether we have any real/equal/no roots where real is >0 (more than 1 root) equal =0 and 1 root and no is <0 with no answers.

You are most certainly correct.

I am suddenly confused again. I got it before what you said in your most recent reply. You say the extra bit of completing the square is the y coordinate, so the value of x will be -2.

You say that the value of x will be -2, which I get because that is what makes the bracket (x-2) equal to 0. You then say you need the y coordinate, which is the "4-c, the extra bit on the end". What do you mean it is the extra bit on the end? Extra bit on the end of where?

These were the steps in the mark scheme:


When you say the 4-c is the extra bit on the end, do you mean on the line where it says: (x+2)^2 = 4-c? That isn't a bit on the end though, it's on the other side of the equation to the (x+2)^2? Can you explain what you mean by this please, I am really confused by this.

And why in this case, is the discriminant just the "4-c" on the RHS of the equation, when the discriminant is normally 4*a*c? Can you explain that too please?

I understood the bit about finding b^2-4ac for the discriminant, setting b^2-4ac>0 and solving it, but your post about the y coordinate being "the extra bit on the end", the end of where? And my second question is why is the discriminant normally b^2-4ac, when in this case, the discriminant is just 4-c? How does that work?

Thanks

Sorry bad wording from me
$(x+2)^2 -4+c=0$

c-4 is the "extra bit" which is added to make the equation equal again to what it originally was.





No the discriminant is $b^2 -4ac$. You have used the discriminant to get an answer of 4-c>0


@notnek need some clarification on how to use completing the square to find a missing constant

Basically, I get how to work out that c<4 by substituting a, b, c from the equation into b^2-4ac to find the discriminant, and working from there.

However, I do not understand how on earth you use the completing the square of (x+2)^2 = 4-c, I get that part. After that though, the mark scheme goes straight to 4-c>0. HOW?! I get that in this instance, b^2-4ac, the discriminant, must be >0. But how do you use the discriminant in this instance to get an answer of 4-c>0? Can you explain that please?

I don't know bro ;-; i'm sorry i just don't understand. On the bright side i've never needed to use completing the square to get an answer. However it's good to know. Also i'm sure we don't use the discriminant for the completing the square method??? Maybe i'm wrong.

@RDKGames and @Zacken need some clarification on how to use completing the square to find a missing constant

Sorry bad wording from me
$(x+2)^2 -4+c=0$

c-4 is the "extra bit" which is added to make the equation equal again to what it originally was.





No the discriminant is $b^2 -4ac$. You have used the discriminant to get an answer of 4-c>0


@notnek need some clarification on how to use completing the square to find a missing constant

No the discriminant is $b^2 -4ac$. You have used the discriminant to get an answer of 4-c>0

Basically, I get how to work out that c<4 by substituting a, b, c from the equation into b^2-4ac to find the discriminant, and working from there.

However, I do not understand how on earth you use the completing the square of (x+2)^2 = 4-c, I get that part. After that though, the mark scheme goes straight to 4-c>0. HOW?! I get that in this instance, b^2-4ac, the discriminant, must be >0. But how do you use the discriminant in this instance to get an answer of 4-c>0? Can you explain that please?

The question is:
"Given that the equation x^2 + 4x + c =0 has unequal real roots, find the range of possible values of c"

From earlier on in the question, I know that x^2 + 4x + c = (x+2)^2 -4 + c =0

I get the (x+2)^2 -4 + c =0 because the questions earlier stages involved completing the square.

However, in the solution, (x+2)^2 -4 + c =0 then goes to (x+2)^2 = 4-c

But it then goes to 4-c > 0.

What happened to the (x+2)^2? Where did that go? And why is it >0?

Thanks

Consider the shape of (x+2)^2 it has a repeated solution so for there to be 2 real solutions you need a negative vertical translation so c-4 <0 which gives c <4

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You have $(x+2)^2=4-c$, so we have $x+2=\pm \sqrt{4-c}$, now tell me how you would solve this if 4-c was a negative number ie. 4-c<0.