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    Find the values of k for which the line y = x + 2 meets the curve y2 + (x + k)2 = 2. ( Answer: 0=/<k</= 4 ). can anyone walk me through the steps, please? TA
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    Square the first equation and substitute into y^2+(x+k)^2=2 to eliminate y.

    As the two lines meet, we know that the resulting equation in x must have real roots. Use this to form a quadratic in k and solve to get two solutions. These will be the inequalities given in the mark scheme. If you're still stuck, post what you have and I'll try and help.
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    (Original post by Damask-)
    Square the first equation and substitute into y^2+(x+k)^2=2 to eliminate y.

    As the two lines meet, we know that the resulting equation in x must have real roots. Use this to form a quadratic in k and solve to get two solutions. These will be the inequalities given in the mark scheme. If you're still stuck, post what you have and I'll try and help.
    ah cheers. yeah still a little stuck-- i end up with (after your instructions):
    2x^2+ k^2 + (4+2k)x+2=0

    how am i meant to solve for k now?
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    (Original post by zomgleh)
    ah cheers. yeah still a little stuck-- i end up with (after your instructions):
    2x^2+ k^2 + (4+2k)x+2=0

    how am i meant to solve for k now?

    So, you've gort a quadratic in x where a=2, b=(4+2k), c=k^2+2.
    What has to be true of the discriminant for this to have real roots?
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    (Original post by Damask-)
    So, you've gort a quadratic in x where a=2, b=(4+2k), c=k^2+2.
    What has to be true of the discriminant for this to have real roots?
    aah ok got it, cheers.

    can you also help me out with this please--

    find a vector in R2 in the direction of the line 2x+7y+3=0 (consider its slope)

    I got the slope as -2/7x and apparently the answer is (-7, 2). I don't get why though?
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    also would you happen to know the taylor approx about 0 for this function--
    (1+x)^α(alpha)?
    the answer is=
    1+αx+α(α-1)x^2/2+ α(α-1)(α-2)x^3/3!+...
 
 
 
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