Can X^3 Equal sqrt{81} with MathMagic Lite?

[mathdad]

Is the following correct?

 

[kaliprasad]

Is the following correct?

yes

 

[MarkFL]

You're actually missing two roots of the equation:

\(\displaystyle x^3=\sqrt{81}\)

Write as difference of cubes:

\(\displaystyle x^3-\left(3^{\frac{2}{3}}\right)^3=0\)

Factor:

\(\displaystyle \left(x-3^{\frac{2}{3}}\right)\left(x^2+3^{\frac{2}{3}}x+3^{\frac{4}{3}}\right)=0\)

The quadratic factor will yield the two complex conjugate roots. ;)

 

[mathdad]

You're actually missing two roots of the equation:

\(\displaystyle x^3=\sqrt{81}\)

Write as difference of cubes:

\(\displaystyle x^3-\left(3^{\frac{2}{3}}\right)^3=0\)

Factor:

\(\displaystyle \left(x-3^{\frac{2}{3}}\right)\left(x^2+3^{\frac{2}{3}}x+3^{\frac{4}{3}}\right)=0\)

The quadratic factor will yield the two complex conjugate roots. ;)

What do you think of MathMagic Lite? Cool, right?

1. Are you saying that my answer is wrong?

2. How badly was your area hit by Irma?

 

[MarkFL]

What do you think of MathMagic Lite? Cool, right?

I much prefer $\LaTeX$...it looks a lot better, can be quoted, and seen in the "Topic Review" element. You can fix the last issue by attaching the images inline, but $\LaTeX$ is still the far superior solution for posting math expressions, IMHO. :)

1. Are you saying that my answer is wrong?

No, I'm just saying there are two other values of $x$ that satisfy the given equation. Now, the answer you gave is real, and the other two solutions are complex, so unless the problem states only to look for real solutions, then we must give all 3.

2. How badly was your area hit by Irma?

It was hit pretty hard, but not as hard as pretty much all points to the south of me. :)

 

[mathdad]

I will continue using MathMagic Lite. It is easy and less stressful to use.