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Find the possible values of K if y=2x+K meets y=x^2-2x-7 in two distinct real points?



thankyou in advance

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Original post by monisj1
Find the possible values of K if y=2x+K meets y=x^2-2x-7 in two distinct real points?



thankyou in advance


Take the first equation from the second. This will give you a quadratic equation, when does then equation have two roots?
Reply 2
Original post by Underthecandle
Take the first equation from the second. This will give you a quadratic equation, when does then equation have two roots?


iv done

2x+K=x^2-2x-7

then

0=x^2-4x-7-K

is that right if so what do i do next?
Original post by monisj1
iv done

2x+K=x^2-2x-7

then

0=x^2-4x-7-K

is that right if so what do i do next?


Yeah that's right, now you have to use the discriminant to find when that has two roots
Reply 4
Original post by Underthecandle
Yeah that's right, now you have to use the discriminant to find when that has two roots


so b^2-4ac>0?

16-4(-7-K)?
Original post by monisj1
so b^2-4ac>0?

16-4(-7-K)?


Yeah, that will give you you values of k
Reply 6
Original post by Underthecandle
Yeah, that will give you you values of k



then thats it?
Original post by monisj1
then thats it?


You have to rearrange 16-4(-7-k)>0 to get k on its own on one side but other than that yeah that's everything
Reply 8
Original post by Underthecandle
You have to rearrange 16-4(-7-k)>0 to get k on its own on one side but other than that yeah that's everything



so i got K>3

but theres supposed to be 2 values?
Original post by monisj1
so i got K>3

but theres supposed to be 2 values?


I don't think k>3 is right how did you get that?

Also why do you think there has to be 2 values?
Reply 10
Original post by Underthecandle
I don't think k>3 is right how did you get that?

Also why do you think there has to be 2 values?



oh is it K>-11


and it sed two distinct real points?
Original post by monisj1
oh is it K>-11


and it sed two distinct real points?


Yeah that's right.

The two distinct pints means that the line y=2x+k needs to cross y=x^2-2-7 twice. This just tells you that the discriminant needs to be bigger than zero and doesn't tell you about how many values of k there is
(edited 9 years ago)
Reply 12
Original post by Underthecandle
Yeah that's right.

The two distinct pints means that the line y=2x+k needs to cross y=x^2-2-7 twice. This just tells you that the discriminant needs to be bigger than zero and doesn't tell you about have many values of k there is



but the discriminant is less than 0 its -11?
Original post by monisj1
but the discriminant is less than 0 its -11?


No the discriminant is 16-4(-7-k) which is bigger than zero as long as k is bigger than -11, (you can try putting some numbers in to check)
Reply 14
Original post by Underthecandle
No the discriminant is 16-4(-7-k) which is bigger than zero as long as k is bigger than -11, (you can try putting some numbers in to check)



you are an amazing person thank you very much!!
Reply 15
Original post by Underthecandle
Take the first equation from the second. This will give you a quadratic equation, when does then equation have two roots?



whats kx times kx?
Original post by monisj1
whats kx times kx?


K^2X^2 or (kx)^2 both are the same thing
Reply 17
Original post by Underthecandle
K^2X^2 or (kx)^2 both are the same thing


im stuck on another one


find the range of values of k for which kx+y=3 meets the x^2+y^2=5 in two distinct points?

so y=3-Kx

therefore

x^2+(3-KX)^2-5=0


so

x^2+k^2x^2-6kx+9-5=0


x^2+k^2x^2-6kx=+4=0



so discrminant is b^2-4ac>0



36k^2x^2-4(x^2+k^2x^2)(4)

36k^2x^2-16x-16k^2x^2

simplifys to


16k^2x^2-16x


im stuck from there?
Original post by monisj1
im stuck on another one


find the range of values of k for which kx+y=3 meets the x^2+y^2=5 in two distinct points?

so y=3-Kx

therefore

x^2+(3-KX)^2-5=0


so

x^2+k^2x^2-6kx+9-5=0


x^2+k^2x^2-6kx=+4=0



so discrminant is b^2-4ac>0



36k^2x^2-4(x^2+k^2x^2)(4)

36k^2x^2-16x-16k^2x^2

simplifys to


16k^2x^2-16x


im stuck from there?


Your discriminant shouldn't have any x in.

If you add k^2x^2 and x^2 you get (k^2 +1)x^2 too
Reply 19
Original post by Underthecandle
Your discriminant shouldn't have any x in.

If you add k^2x^2 and x^2 you get (k^2 +1)x^2 too



i am confused?

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