# 2004_Question2

(a)

Write the equation as:

x^2 - 2x - 1 - k = 0

(i) We have no real solutions if:

2^2 - 4 \times (-1 - k) < 0

That is:

1 < -1 - k \implies k + 1 < -1 \implies k < -2

(ii) We have one real solution if:

2^2 - 4(-1 - k) = 0

That is, if k = -2.

(iii) We have two real solution if:

2^2 - 4(-1- k) = 0

That is, if k > -2.

(b) We have:

\begin{align*}(x^2 - 2x - 1)^2 & = ((x^2 - 1) - 2x)^2 \\ & = (x^2 - 1)^2 - 4x(x^2 - 1) + 4x^2 \\ & = x^4 - 2x^2 + 1 - 4x^3 + 4x + 4x^2 \\ & = x^4 - 4x^3 + 2x^2 + 4x + 1\end{align*}

© Note that we necessarily have h \ge 0. We can rewrite the equation as:

(x^2 - 2x - 1)^2 = h = (\sqrt h)^2

So either x^2 - 2x - 1 = \sqrt h or x^2 - 2x - 1 = -\sqrt h.

We could have one of the following cases:

• One equation has two real solutions and the other has none.
• Both equations have the same two real solutions
• Both equations have one real solution, both distinct

In the first case, since \sqrt h \ge 0, (and we require k < -2 for no real solutions) the latter equation must be the equation with no real solutions. This occurs iff -\sqrt h < -2, ie. h > 4.

In the second case the quadratics must be multiples of eachother, (consider the fundamental theorem of algebra) which happens if \sqrt h = -\sqrt h, ie. h = 0.

The third case is impossible, since \sqrt h \ge 0, the equation x^2 - 2x - 1 = \sqrt h cannot have only a single real solution. (since we would require \sqrt h = -2)

So either h = 0, h > 4. (this is in fact an iff)

As in the third case of the above argument - we would require x^2 - 2x - 1 = \sqrt h and x^2 - 2x - 1 = -\sqrt h to share the same single real solution. Either note that they can only share (all their) solutions if \sqrt h = 0, and in this case there are two real solutions or noting that we’ve already shown that x^2 - 2x - 1 = \sqrt h cannot have only a single real solution.