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September 23, 2019, 06:16:44 am

Author Topic: Methods - Help with question  (Read 230 times)  Share 

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Travishobo

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Methods - Help with question
« on: February 05, 2019, 09:29:46 pm »
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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!

AlphaZero

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Re: Methods - Help with question
« Reply #1 on: February 05, 2019, 11:42:19 pm »
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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:
> two solutions for \(x\) if \(\Delta>0\),
> one solution for \(x\) if \(\Delta=0\), and
> 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.
_______________________________________________________________________________

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 \(k\in\mathbb{R}\).
« Last Edit: February 05, 2019, 11:50:32 pm by dantraicos »
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne


Travishobo

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Re: Methods - Help with question
« Reply #2 on: February 06, 2019, 10:58:37 pm »
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Thank you dantraicos!