Hello sym0110,

for delta, you can use the \Delta command : $\displaystyle \Delta$

For the second question, you must show that for any $\displaystyle k$, the equation $\displaystyle x + 1 = 4x^2 - 2kx + k$ has at least one real solution in $\displaystyle x$. Check the discriminant of this quadratic equation to see if it can actually have real solutions (negative discriminant means no real solution). Can you follow up ?

Take your quadratic equation : $\displaystyle 4x^2 - 2kx + k - x - 1 = 0$.

This can be factorized as : $\displaystyle 4x^2 - (2k + 1)x + (k - 1) = 0$.

Take the discriminant : $\displaystyle \Delta = b^2 - 4ac = (2k + 1)^2 - 4 \times 4 \times (k - 1)$.

We then have : $\displaystyle \Delta = (2k + 1)^2 - 16(k - 1)$.

We want to prove that $\displaystyle \Delta > 0$, so $\displaystyle (2k + 1)^2 - 16(k - 1) > 0$.

That is : $\displaystyle (2k)^2 + 4k + 1^2 - 16k + 16 > 0$

Simplify further : $\displaystyle 4k^2 + 4k + 1 - 16k + 16 > 0$

Keep going : $\displaystyle 4k^2 - 12k + 17 > 0$

Is this correct ? Let us check the discriminant of this new quadratic :

$\displaystyle \Delta ' = 12^2 - 4 \times 4 \times 17 = 144 - 272 = -128 \ (< 0)$.

If the discriminant is negative, it means the equation has no real roots and thus all values are greater than zero (this is because the $\displaystyle a$ value of the quadratic equation is positive. If it was negative, then all values would be less than zero. Graph some equations to see what I mean). Thus we have proved that :

$\displaystyle 4k^2 - 12k + 17 > 0$

And therefore we have proved that for any real $\displaystyle k$, the line intersects the curve (finish the conclusion properly).