Thread: Please help with solving for x when a 4th degree equation is involved.

I have mixed ideas as to how to solve this equation:

2x^4-8x^2+12=0

I found something in my book that says: to solve a 4th degree equation use substitution to rewrite the equation as a quadratic equation. So you'd let u = x^2. However there is a coefficient here, so I wasn't sure how I should attempt to factor it.

I thought about:

(2x^2 - 6) (x^2-2) = 0 but my answers didn't check. I had (2, -2).

So, I think I'm going to need help (once again). Please provide any formulas or rules that are applicable to equations of this type. Thanks in advance!

Originally Posted by lilrhino

I have mixed ideas as to how to solve this equation:

2x^4-8x^2+12=0

I found something in my book that says: to solve a 4th degree equation use substitution to rewrite the equation as a quadratic equation. So you'd let u = x^2. However there is a coefficient here, so I wasn't sure how I should attempt to factor it.

I thought about:

(2x^2 - 6) (x^2-2) = 0 but my answers didn't check. I had (2, -2).

So, I think I'm going to need help (once again). Please provide any formulas or rules that are applicable to equations of this type. Thanks in advance!

Only even powers of x appear in the equation, so I agree with making the substitution u = x^2.
2u^2 - 8u + 12 = 0

The reason your factoring doesn't give you the right answers is because you factored wrong. Try this first: each term has a common factor of 2, so:
2u^2 - 8u + 12 = 0

2(u^2 - 4u + 6) = 0

u^2 - 4u + 6 = 0

It should be relatively easy to see that this doesn't factor. So use the quadratic formula:

u = [4 (+/-) sqrt{16 - 4*1*6}]/(2*1)

u = [4 (+/-) sqrt{-8}]/2

u = 2 (+/-) I*sqrt{2}, where I^2 = -1

Now, u = x^2 so

x^2 = 2 (+/-) I*sqrt{2}

So
x = (+/-)sqrt{2 (+/-) I*sqrt{2}} <-- Where the (+/-) signs are not related.
(Note that there are 4 solutions and all of them are complex.)

This will do for an answer, but it is not simplified.

-Dan

Originally Posted by topsquark

Only even powers of x appear in the equation, so I agree with making the substitution u = x^2.
2u^2 - 8u + 12 = 0

The reason your factoring doesn't give you the right answers is because you factored wrong. Try this first: each term has a common factor of 2, so:
2u^2 - 8u + 12 = 0

2(u^2 - 4u + 6) = 0

u^2 - 4u + 6 = 0

It should be relatively easy to see that this doesn't factor. So use the quadratic formula:

u = [4 (+/-) sqrt{16 - 4*1*6}]/(2*1)

u = [4 (+/-) sqrt{-8}]/2

u = 2 (+/-) I*sqrt{2}, where I^2 = -1

Now, u = x^2 so

x^2 = 2 (+/-) I*sqrt{2}

So
x = (+/-)sqrt{2 (+/-) I*sqrt{2}} <-- Where the (+/-) signs are not related.
(Note that there are 4 solutions and all of them are complex.)

This will do for an answer, but it is not simplified.

-Dan

Thanks for your response Dan. Gosh, this is complicated. I think I get stuck when something doesn't factor. So, I could've used the quadratic formula since it didn't factor .

What do you mean exactly when you say that the answer is not simplified? Would an answer such as this be acceptable on a test?

Originally Posted by lilrhino

Thanks for your response Dan. Gosh, this is complicated. I think I get stuck when something doesn't factor. So, I could've used the quadratic formula since it didn't factor .

What do you mean exactly when you say that the answer is not simplified? Would an answer such as this be acceptable on a test?

Probably. Let me show you how to simplify it, then you can decide for yourself if you would shoot your professor for making you simplify it.

Each of the four solutions is a point on the Argand plane, that is to say of the form x + I*y = r*e^{I*t} = r*cos(t) + I*r*sin(t) (where I'm using t as shorthand for "theta.") We want to express 2 (+/-) I*sqrt{2} in terms of r and t, then we may easily take the square root.

So. Let's take the "+" solution first.
2 + I*sqrt{2} = r*cos(t) + I*r*sin(t)

So
2 = r*cos(t)
sqrt{2} = r*sin(t)

Dividing the two gives:
[r*sin(t)]/[r*cos(t)] = sqrt{2}/2

tan(t) = sqrt{2}/2 ==> t = 0.61548 rad <-- There is no exact solution for this.

Thus:
r = 2/cos(t) = 2.44949

Thus
2 + I*sqrt{2} = 2.44949*cos(0.61548) + I*2.44949*sin(0.61548) = 2.44949*e^{0.61548I}

So
x = (+/-)sqrt{2 + I*sqrt{2}} = (+/-)sqrt{2.44949*e^{0.61548I}}

x = (+/-)sqrt{2.44949}*e^{0.61548I/2} = 1.56508*e^{0.30774I}

x = (+/-)1.56508*e^{0.30774I} = (+/-)[1.49156 + 0.474073I] <-- Either form should be suitable.

I'll let you do the two x solutions with the 2 - I*sqrt{2}.

-Dan

Originally Posted by topsquark

Probably. Let me show you how to simplify it, then you can decide for yourself if you would shoot your professor for making you simplify it.

Each of the four solutions is a point on the Argand plane, that is to say of the form x + I*y = r*e^{I*t} = r*cos(t) + I*r*sin(t) (where I'm using t as shorthand for "theta.") We want to express 2 (+/-) I*sqrt{2} in terms of r and t, then we may easily take the square root.

So. Let's take the "+" solution first.
2 + I*sqrt{2} = r*cos(t) + I*r*sin(t)

So
2 = r*cos(t)
sqrt{2} = r*sin(t)

Dividing the two gives:
[r*sin(t)]/[r*cos(t)] = sqrt{2}/2

tan(t) = sqrt{2}/2 ==> t = 0.61548 rad <-- There is no exact solution for this.

Thus:
r = 2/cos(t) = 2.44949

Thus
2 + I*sqrt{2} = 2.44949*cos(0.61548) + I*2.44949*sin(0.61548) = 2.44949*e^{0.61548I}

So
x = (+/-)sqrt{2 + I*sqrt{2}} = (+/-)sqrt{2.44949*e^{0.61548I}}

x = (+/-)sqrt{2.44949}*e^{0.61548I/2} = 1.56508*e^{0.30774I}

x = (+/-)1.56508*e^{0.30774I} = (+/-)[1.49156 + 0.474073I] <-- Either form should be suitable.

I'll let you do the two x solutions with the 2 - I*sqrt{2}.

-Dan

I'll give it a shot Dan...phew this one makes me want to get out another can of Mountain Dew.

It'll take me a while, my brain doesn't work as fast as I'd like it to. Thanks.