Original post by B_9710

Substitute y=2x into equation of the circle. This forms a quadratic in x. Now if the line and circle do not intersect then how many roots should this equation have?

Spoiler

Please elucidate

Original post by Grade9girl

Please elucidate

Note that this thread is a year old

Substituting the equation of the line into the equation of the circle gives you a quadratic equation in x. If the line does not intersect the circle, then the equation must have no real roots. Which property of a quadratic equation tells you about the types of roots it has?

Original post by Grade9girl

Please elucidate

hi, i just did this question on integral as a homework ðŸ˜‚ and i was stuck on it as well so i searched the question up and found ur thread. I know how to do it now so i thought I'd reply to ur question. I don't really know if you need it anymore tho

so..

sub y=2x into the equation of the circle:

(x-2)^2+(2x-1)^2=d...

that becomes..5x^2-8x+5=d

Now basically what we have is a quadratic and since the circle doesn't intrsect the line y=2x we use b^2-4ac<0

but before u do this you need to take d to the other side so that c becomes (5-d):

5x^2-8x+5=d

5x^2-8x+(5-d)=0

now substitute ur a b and c values into b^2-4ac<0:

(-8)^2-4*5*(5-d)<0

that becomes... 64-100+20d<0

rearrange and you get... d<1.8

Original post by silverbullet786

hi, i just did this question on integral as a homework ðŸ˜‚ and i was stuck on it as well so i searched the question up and found ur thread. I know how to do it now so i thought I'd reply to ur question. I don't really know if you need it anymore tho

so..

sub y=2x into the equation of the circle:

(x-2)^2+(2x-1)^2=d...

that becomes..5x^2-8x+5=d

Now basically what we have is a quadratic and since the circle doesn't intrsect the line y=2x we use b^2-4ac<0

but before u do this you need to take d to the other side so that c becomes (5-d):

5x^2-8x+5=d

5x^2-8x+(5-d)=0

now substitute ur a b and c values into b^2-4ac<0:

(-8)^2-4*5*(5-d)<0

that becomes... 64-100+20d<0

rearrange and you get... d<1.8

so..

sub y=2x into the equation of the circle:

(x-2)^2+(2x-1)^2=d...

that becomes..5x^2-8x+5=d

Now basically what we have is a quadratic and since the circle doesn't intrsect the line y=2x we use b^2-4ac<0

but before u do this you need to take d to the other side so that c becomes (5-d):

5x^2-8x+5=d

5x^2-8x+(5-d)=0

now substitute ur a b and c values into b^2-4ac<0:

(-8)^2-4*5*(5-d)<0

that becomes... 64-100+20d<0

rearrange and you get... d<1.8

thank you!!! idk if they other person still needed it but i certainly did haha

i just did the same integral test and i entered the answer as a decimal instead of fraction and missed one mark, i wouldâ€™ve gotten 100%, what does it matter anyway, they both mean the same thing!! ðŸ˜¤ðŸ˜¤ðŸ˜¤ lmao iâ€™m so annoyed for no reason

Original post by weran

thank you!!! idk if they other person still needed it but i certainly did haha

i just did the same integral test and i entered the answer as a decimal instead of fraction and missed one mark, i wouldâ€™ve gotten 100%, what does it matter anyway, they both mean the same thing!! ðŸ˜¤ðŸ˜¤ðŸ˜¤ lmao iâ€™m so annoyed for no reason

i just did the same integral test and i entered the answer as a decimal instead of fraction and missed one mark, i wouldâ€™ve gotten 100%, what does it matter anyway, they both mean the same thing!! ðŸ˜¤ðŸ˜¤ðŸ˜¤ lmao iâ€™m so annoyed for no reason

ðŸ˜‚ðŸ˜‚ lol definitely made that mistake before!!! dont worry i only got 1/3!!! I suppose that means you take A-level maths. Dont wprry were in the same boat ðŸ˜…ðŸ˜…ðŸ˜…

(edited 3 years ago)

thanks, that really helped me understand how to do it.

Original post by Zahra575!

hi, i just did this question on integral as a homework ðŸ˜‚ and i was stuck on it as well so i searched the question up and found ur thread. I know how to do it now so i thought I'd reply to ur question. I don't really know if you need it anymore tho

so..

sub y=2x into the equation of the circle:

(x-2)^2+(2x-1)^2=d...

that becomes..5x^2-8x+5=d

Now basically what we have is a quadratic and since the circle doesn't intrsect the line y=2x we use b^2-4ac<0

but before u do this you need to take d to the other side so that c becomes (5-d):

5x^2-8x+5=d

5x^2-8x+(5-d)=0

now substitute ur a b and c values into b^2-4ac<0:

(-8)^2-4*5*(5-d)<0

that becomes... 64-100+20d<0

rearrange and you get... d<1.8

so..

sub y=2x into the equation of the circle:

(x-2)^2+(2x-1)^2=d...

that becomes..5x^2-8x+5=d

Now basically what we have is a quadratic and since the circle doesn't intrsect the line y=2x we use b^2-4ac<0

but before u do this you need to take d to the other side so that c becomes (5-d):

5x^2-8x+5=d

5x^2-8x+(5-d)=0

now substitute ur a b and c values into b^2-4ac<0:

(-8)^2-4*5*(5-d)<0

that becomes... 64-100+20d<0

rearrange and you get... d<1.8

I know we shouldn't revive old threads (to admin, sorry about commenting even more), but this is proof maths questions are timeless lol

Original post by 1-1AR31-1

thanks, that really helped me understand how to do it.

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