July 07, 2020, 11:57:10 am

### AuthorTopic: Methods - Help with question  (Read 323 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#### Travishobo

• Posts: 16
• Respect: 0
##### Methods - Help with question
« on: February 05, 2019, 09:29:46 pm »
0
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.

#### AlphaZero

• MOTM: DEC 18
• Forum Obsessive
• Posts: 348
• $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$
• Respect: +154
##### Re: Methods - Help with question
« Reply #1 on: February 05, 2019, 11:42:19 pm »
+3
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.

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\\
&>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 Mathematical Sciences Diploma, University of Melbourne