I equated derivatives to get kx=1/4(x-k)^1/2, then substituted back into the originals to get x in terms of k. Then again , substitution to get the original equations in terms of x only
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Similar to what I did, though I squared up both equations and noted its where d y^2 / dx = 1 (second equation), which gives k (= 1/(2x^(3/2)) directly from the first equation. Then as you say sub to get x.
aug 23, 2025, 12:26 pm • 0 0 • view
Yes, with this problem I wasn't sure whether it would be quicker to do the repeated root approach or the equal derivatives approach. I find that sometimes one route is clearly better than the other, but that wasn't the case here.
aug 23, 2025, 9:48 am • 2 0 • view
Can I ask please, for the repeated root approach, if the LHS is k^2x^4-x+k, what is the RHS?
aug 23, 2025, 11:46 am • 1 0 • view
The LHS you have must factor as (x-c)²(k²x²+dx+e) for some c, as there's a double root at the intersection point. You can expand and then compare coefficients.
aug 23, 2025, 12:30 pm • 1 0 • view
Thankyou
aug 23, 2025, 12:42 pm • 1 0 • view
With some flexibility about what you do with the k²... 😉
aug 23, 2025, 12:43 pm • 1 0 • view
Indeed - I went for (kx-c)²...
aug 23, 2025, 1:00 pm • 2 0 • view
I factorised the k² out the front 😀
aug 23, 2025, 1:20 pm • 1 0 • view