Author Topic: Quadratics! (Read 1354 times)

zara · « **on:** September 28, 2009, 10:26:58 pm »

Find the value of k such tht the st. line y=2x+k meets the curve x²+2xy+2y²=5.

i started buh got stuck in between....

answer urgently anyone if possible!
thanks!

nid404 · « **Reply #1 on:** September 29, 2009, 08:07:03 am »

what's the answer?

Ghost Of Highbury · « **Reply #2 on:** September 29, 2009, 09:24:04 am »

i'm getting this answer..

k<8.06

is that correct??

nid404 · « **Reply #3 on:** September 29, 2009, 09:29:04 am »

post your steps plz

Ghost Of Highbury · « **Reply #4 on:** September 29, 2009, 09:33:15 am »

lemme try again...i
ll post soon

astarmathsandphysics · « **Reply #5 on:** September 29, 2009, 01:16:34 pm »

Quote from: Z.J. on September 28, 2009, 10:26:58 pm

Find the value of k such tht the st. line y=2x+k meets the curve x²+2xy+2y²=5.

i started buh got stuck in between....

answer urgently anyone if possible!
thanks!

$x^2+2xy+y^2=5$

Substitute y=2x+k to get $x^2+2x(2x+k)+2(2x+k)^2=5$ so $x^2+4x^2+2xk+8x^2+8xk+2k^2-5=0$
so $13x^2+10kx+2k^2-5=0$

Since the curve meets the line at only one point the discriminant= $b^2-4ac=0$ where a=13, b=10k and c= $2k^2-5$

We get $(10k)^2-4*13*(2k^2-5)=0$ which simplifies to -4k^2+260=0 so k= $\pm sqrt(65)$

Ghost Of Highbury · « **Reply #6 on:** September 29, 2009, 01:19:06 pm »

yuppiee!!

my answer was correct...except the > sign

astarmathsandphysics · « **Reply #7 on:** September 29, 2009, 01:19:54 pm »

I saw this yeasterday but forgot and only noticed it again this morning.

zara · « **Reply #8 on:** September 29, 2009, 06:19:26 pm »

thanks astar n adi!
i figured out my mistake later...=]

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Author Topic: Quadratics! (Read 1354 times)

zara

Quadratics!

nid404

Re: Quadratics!

Ghost Of Highbury

Re: Quadratics!

nid404

Re: Quadratics!

Ghost Of Highbury

Re: Quadratics!

astarmathsandphysics

Re: Quadratics!

Ghost Of Highbury

Re: Quadratics!

astarmathsandphysics

Re: Quadratics!

zara

Re: Quadratics!