is this a valid way to solve this equation?

[allegansveritatem]

The equation:


My solution:


I just kept factoring removing factors once I had zeroed them. I don't recall ever doing it this way before. Is it right? Is it the long way around? Is there a quicker way? I know there is a formula for solving 4th degree equations but I've read that it is a cure that is worse than the disease. No?

 

[tkhunny]

There is a process / formula for quartic polynomials. It is more of an existential thing, rather than a helpful exercise.

If you are lucky, you will be able to factor.

Noting a factor of 'x', one might have noted that solution, first, and reduced to a cubic polynomial.

There is a process / formula for cubic polynomials. It is more of an existential thing, rather than a helpful exercise.

If you are lucky, you will be able to factor.

Where did the "24" come from? That's no good.

 

[ksdhart2]

Your technique is perfectly valid, yes. Additionally, we can verify it produced the correct answers by multiplying the polynomial back together:

\(\displaystyle \left(x - 0 \right) \left(x + \frac{5}{2} \right) \left(x - \frac{\sqrt{6}}{2} \right) \left(x + \frac{\sqrt{6}}{2} \right)\)

\(\displaystyle = x \left(x + \frac{5}{2} \right) \left(x^2 - \frac{3}{2} \right)\)

\(\displaystyle = x \left(x^3 + \frac{5x^2}{2} - \frac{3x}{2} - \frac{15}{4} \right)\)

\(\displaystyle = x^4 + \frac{5x^3}{2} - \frac{3x^2}{2} - \frac{15x}{4}\)

\(\displaystyle = 4x^4 + 10x^3 - 6x^2 - 15x\)

As for the quartic formula, a general formula for solving 4th degree polynomials does exist, but you're correct that it's an awful, vile thing. In my opinion, you're almost always best off with some other technique unless you absolutely 100% need exact closed form solutions.

Also:

Where did the "24" come from? That's no good.

The 24 comes from using the quadratic formula on \(2x^2 - 3 = 0\). We have A = 2, B = 0, and C = -3, so the discriminant is 24.

 

[Steven G]

Your technique is perfectly valid, yes. Additionally, we can verify it produced the correct answers by multiplying the polynomial back together:

\(\displaystyle \left(x - 0 \right) \left(x + \frac{5}{2} \right) \left(x - \frac{\sqrt{6}}{2} \right) \left(x + \frac{\sqrt{6}}{2} \right)\)

\(\displaystyle = x \left(x + \frac{5}{2} \right) \left(x^2 - \frac{3}{2} \right)\)

\(\displaystyle = x \left(x^3 + \frac{5x^2}{2} - \frac{3x}{2} - \frac{15}{4} \right)\)

\(\displaystyle = x^4 + \frac{5x^3}{2} - \frac{3x^2}{2} - \frac{15x}{4}\)

\(\displaystyle = 4x^4 + 10x^3 - 6x^2 - 15x\)

As for the quartic formula, a general formula for solving 4th degree polynomials does exist, but you're correct that it's an awful, vile thing. In my opinion, you're almost always best off with some other technique unless you absolutely 100% need exact closed form solutions.

Also:



The 24 comes from using the quadratic formula on \(2x^2 - 3 = 0\). We have A = 2, B = 0, and C = -3, so the discriminant is 24.

Where did that last equal sign come from?
You should use the quartic formula if absolutely 100% need an exact closed form solution--really??

 

[ksdhart2]

Where did that last equal sign come from?

Ah, right. That's an error on my part. They're not equal, but it's still okay to multiply each term by 4 to "clear" the fractions since we're interested in setting it equal to 0 to find the roots.

You should use the quartic formula if absolutely 100% need an exact closed form solution--really??

Sure. What's wrong with that? If you for some reason need exact closed form solutions to some arbitrary quartic equation, and really good approximations just won't do, the quartic formula seems okay to use. What I was saying though is I wouldn't recommend it in any other situation. I'd only use it if I needed exact solutions and the rational root theorem and factoring by grouping and such had failed.

 

[Steven G]

I have a little problem with your work. You started off with an equation (It had an equal sign), then every other line has no equal sign. You really shoul write the 0s.

As tkhunny pointed out in solving 2x²=3, you should then say x²=3/2 and x = +/-sqrt(3/2) = +/- sqrt(6)/2

 

[Steven G]

Ah, right. That's an error on my part. They're not equal, but it's still okay to multiply each term by 4 to "clear" the fractions since we're interested in setting it equal to 0 to find the roots.



Sure. What's wrong with that? If you for some reason need exact closed form solutions to some arbitrary quartic equation, and really good approximations just won't do, the quartic formula seems okay to use. What I was saying though is I wouldn't recommend it in any other situation. I'd only use it if I needed exact solutions and the rational root theorem and factoring by grouping and such had failed.

Ah, but you did NOT say and the rational root theorem and factoring by grouping and such had failed.
You clearly said that if you need exact solutions then use the quartic formula.

 

[allegansveritatem]

Your technique is perfectly valid, yes. Additionally, we can verify it produced the correct answers by multiplying the polynomial back together:

\(\displaystyle \left(x - 0 \right) \left(x + \frac{5}{2} \right) \left(x - \frac{\sqrt{6}}{2} \right) \left(x + \frac{\sqrt{6}}{2} \right)\)

\(\displaystyle = x \left(x + \frac{5}{2} \right) \left(x^2 - \frac{3}{2} \right)\)

\(\displaystyle = x \left(x^3 + \frac{5x^2}{2} - \frac{3x}{2} - \frac{15}{4} \right)\)

\(\displaystyle = x^4 + \frac{5x^3}{2} - \frac{3x^2}{2} - \frac{15x}{4}\)

\(\displaystyle = 4x^4 + 10x^3 - 6x^2 - 15x\)

As for the quartic formula, a general formula for solving 4th degree polynomials does exist, but you're correct that it's an awful, vile thing. In my opinion, you're almost always best off with some other technique unless you absolutely 100% need exact closed form solutions.

Also:



The 24 comes from using the quadratic formula on \(2x^2 - 3 = 0\). We have A = 2, B = 0, and C = -3, so the discriminant is 24.

well, the 2x^2-3 got turned into a quadratic equation like this: 2x^2+0x-3=0 and then I used the quadratic formula.

 

[allegansveritatem]

Ah, right. That's an error on my part. They're not equal, but it's still okay to multiply each term by 4 to "clear" the fractions since we're interested in setting it equal to 0 to find the roots.



Sure. What's wrong with that? If you for some reason need exact closed form solutions to some arbitrary quartic equation, and really good approximations just won't do, the quartic formula seems okay to use. What I was saying though is I wouldn't recommend it in any other situation. I'd only use it if I needed exact solutions and the rational root theorem and factoring by grouping and such had failed.

Actually, until I checked the rational root theorem I didn't know a thing about it. I watched a couple of videos on Youtube and got some idea but I will have to do a bunch of problems with it to really learn it.

 

[allegansveritatem]

I have a little problem with your work. You started off with an equation (It had an equal sign), then every other line has no equal sign. You really shoul write the 0s.

As tkhunny pointed out in solving 2x²=3, you should then say x²=3/2 and x = +/-sqrt(3/2) = +/- sqrt(6)/2

Right. I will watch that. I was groping along through this problem and was focused on showing how the factors cancelled out, or should I say "zeroed out"?