Question about a double tangent to a curve.?

y=mx+c meets curve when x⁴+x(mx+c)+(mx+c)²+x = 7

i.e. when f(x) = x⁴ + mx²(1+m) + x(1+c+2cm) + (c²−7) = 0

If line is a double tangent then this equation must have two double roots in x

Hence f(x) = (x−a)²(x−b)² = x⁴ + mx²(1+m) + x(1+c+2cm) + (c²−7) for some a,b

Expanding LHS to x⁴ − 2x³(a+b) + x²(a²+4ab+b²) – 2abx(a+b) + a²b²

And comparing coefficients gives

a+b = 0 … (i)

a²+4ab+b² = m(1+m) … (ii)

−2ab(a+b) = 1+c+2cm … (iii)

a²b² = c²−7 … (iv)

Write (ii) as (a+b)²+2ab = m(1+m) and use (i) to get ab = m(1+m)/2

Using (i) in (iii) gives 1+c+2cm = 0 so c = −1(1+2m)

Put these results in (iv) : m²(1+m)²/4 = 1/(1+2m)² − 7

→ 4m⁶ + 12m⁵ + 13m⁴ + 6m³ + 113m² + 112m + 24 = 0

This has two real roots m ≈ −0.688827382494622, −0.311172617505378

First gives required tangent y = 2.647921045107112679 − 0.6888273824946220000x

I don't think that such a solution exists...

Q1 and Q2... y> 0

y = sqrt(7-x^4+1/4 x^2-x)- 1/2 x

This curve is nearly a line.

certainly there are values for y where

y = -sqrt(7-x^4+1/4 x^2-x)- 1/2 x, but y is not in Q1 or Q2...