I am struggling to understand this question. I believe that the worked solution is also incorrect since the discriminant is not 4 + k(k+1). Shouldn't it be 4+4k^2+4k.

Thanks in advance!

Indeed, the solution is wrong, but the logic they are using is (almost) correct.

Where \(\Delta=b^2-4ac\) and \(a\neq 0\), recall that the equation \(ax^2+bx+c=0\) will have:

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*two* solutions for \(x\) if \(\Delta>0\),

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*one* solution for \(x\) if \(\Delta=0\), and

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*no* solutions for \(x\) if \(\Delta<0\).

So, to prove that the equation \((k+1)x^2-2x-k=0\) always has at least one solution for \(x\) for all \(k\in\mathbb{R}\), it's sufficient to show that \(\Delta=(-2)^2-4(k+1)(-k)\geq 0\) whenever \(k\neq -1\) AND that when \(k=-1\), there is a solution.

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Here is the proof: \begin{align*}\text{For }k\neq -1,\ \ \Delta&=4+4k^2+4k\\

&=4(k^2+k+1)\\

&=4\left[\left(k+\frac12\right)^2-\frac14+1\right]\\

&=4\left(k+\frac12\right)^2+3\\

&\geq 3\quad\quad\qquad\qquad\qquad\qquad \left[\text{since }\left(k+\frac12\right)^2\geq0\right]\\

&>0\end{align*} Also, if \(k=-1\), then we have \(x=\dfrac12\).

Hence, the equation \((k+1)x^2-2x-k=0\) always has a solution for \(x\) for all \(k\in\mathbb{R}\).