# Math Help - Cubic solving problem?

1. ## Cubic solving problem?

Hi all

I have a casio CFX-9850GC PLUS graphic calculator. I tried to solve the cubic: $96x + 128x^3=0$ by inputting:
a= 128, b=0, c=96, d=0

My friend has a casio 9750 (cant remember any letters) and his solved the equation fine by getting 2, yet my calculator gets 0.866i and -0.866i.

Can anyone help me get the right answer at all? Firstly, I don't know what the 'i' bit is and I dont know why it isnt getting the correct answer?

Im new to graphic calculators, but this is over my head, and I've looked in the instruction manual but havent had any success.

I need this calculator my my Core 2 AS level maths exam in a few weeks, and I need to be completely sure that it is working properly and efficiently.

Many thanks!

2. Originally Posted by fishkeeper
Hi all

I have a casio CFX-9850GC PLUS graphic calculator. I tried to solve the cubic: $96x + 128x^3=0$ by inputting:
a= 128, b=0, c=96, d=0

My friend has a casio 9750 (cant remember any letters) and his solved the equation fine by getting 2, yet my calculator gets 0.866i and -0.866i.
Your calculator is right: This equation has only the real solution $x_1=0$ and the two imaginary solutions $x_{2,3}=\pm \frac{\sqrt{3}}{2}i\approx \pm 0.866i$.

3. Thanks for your reply, but, how come following the following mark scheme, the answer of the stationary point is (0,2)?

Because stationary points are found by factorising, and my friends calculator factorised the cubic and got the answer that mark scheme said.

Im confused. Sorry, excuse my ignorance please

p.s here is the question and mark scheme:

hhhh-1.jpg picture by fishkeeperpro - Photobucket

The photo embedder isnt working?

4. Originally Posted by fishkeeper
Thanks for your reply, but, how come following the following mark scheme, the answer of the stationary point is (0,2)?

Because stationary points are found by factorising, and my friends calculator factorised the cubic and got the answer that mark scheme said.

Im confused. Sorry, excuse my ignorance please

p.s here is the question and mark scheme:

hhhh-1.jpg picture by fishkeeperpro - Photobucket

The photo embedder isnt working?

The stationary point $(0, 2)$ is of the form $(x, f(x))$.

You gave us the cubic $f'(x)$. We found that the only real solution to $f'(x)=0$ is given by $x = 0$. So the stationary point is at $(0, f(0))$.

As to how to get your calculator to give you real solutions rather than complex, I'm not sure. There is probably some setting you can adjust.

By the way, this problem easily reduces to solving a quadratic by simply factoring out an $x$ at the beginning.