Solving quintic equation

[Qwertyuiop[]]

Hi, i have this equation of degree 5. The equation is [math](x^5 - 4x^3)^2 =0[/math]
I did it like this , i just need to know if i did it right. So i factor out x^3 and we get: [math](x^3(x^2-4))^2 = 0[/math]I took the square root of both sides and set equal to 0 and got x=0 , x=2 , x=-2.

The other way is to expand but i get an equation of degree 10 and it gets a little complicated. How would you solve this, by factoring out the x^6 ?
This is the expanded form : x^10 - 8(x^8) + 16(x^6)= 0

 

[AvgStudent]

How would you solve this, by factoring out the x^6 ?

[imath](x^5 - 4x^3)^2 =x^6(x^2-4)^2=0[/imath]

It follows that [imath]x^6=0[/imath] or [imath](x^2-4)^2=0 \implies x=0,\pm 2.[/imath]

Hope this helps

 

[blamocur]

Hi, i have this equation of degree 5. The equation is [math](x^5 - 4x^3)^2 =0[/math]
I did it like this , i just need to know if i did it right. So i factor out x^3 and we get: [math](x^3(x^2-4))^2 = 0[/math]I took the square root of both sides and set equal to 0 and got x=0 , x=2 , x=-2.

The other way is to expand but i get an equation of degree 10 and it gets a little complicated. How would you solve this, by factoring out the x^6 ?
This is the expanded form : x^10 - 8(x^8) + 16(x^6)= 0

I don't see any problems with your solution.

 

[JeffM]

There is nothing wrong with what you did.

I might do it as follows

[math]\{x^3(x^2 - 4)\}^2 = 0 \implies \{x^3(x - 2)(x + 2)\}^2 = 0 \implies \\ x^3(x - 2)(x + 2) = \sqrt{0} = 0 \implies \text { by the zero-product property that} \\ x = 0, \ x = 2, \text { or } x = - 2.[/math]

 

[Qwertyuiop[]]

My concern is that do i loose solutions when i take the square root ?

 

[Qwertyuiop[]]

There is nothing wrong with what you did.

I might do it as follows

[math]\{x^3(x^2 - 4)\}^2 = 0 \implies \{x^3(x - 2)(x + 2)\}^2 = 0 \implies \\ x^3(x - 2)(x + 2) = \sqrt{0} = 0 \implies \text { by the zero-product property that} \\ x = 0, \ x = 2, \text { or } x = - 2.[/math]

oh yes, the difference of two squares, i missed that . Just one thing, by taking square root of both sides, can we loose solutions?

 

[JeffM]

There is only one square root of zero so you lose nothing and add nothing with [imath]\sqrt{0}[/imath].

 

[Qwertyuiop[]]

There is only one square root of zero so you lose nothing and add nothing with [imath]\sqrt{0}[/imath].

So if it was let's say 4, then we get 2, but we loose -2 right?

 

[Deleted member 4993]

Hi, i have this equation of degree 5. The equation is [math](x^5 - 4x^3)^2 =0[/math]
I did it like this , i just need to know if i did it right. So i factor out x^3 and we get: [math](x^3(x^2-4))^2 = 0[/math]I took the square root of both sides and set equal to 0 and got x=0 , x=2 , x=-2.

The other way is to expand but i get an equation of degree 10 and it gets a little complicated. How would you solve this, by factoring out the x^6 ?
This is the expanded form : x^10 - 8(x^8) + 16(x^6)= 0

How many roots did you find for x^10 - 8(x^8) + 16(x^6)= 0 ?

 

[JeffM]

So if it was let's say 4, then we get 2, but we loose -2 right?

So if it was let's say 4, then we get 2, but we loose -2 right?

If we are talking about real solutions of

[imath]\{(x^3(x^2 - 4)\}^2 = 4[/imath]. we are in a completely different world. There may be as many as ten or as few as no real solutions.

 

[Deleted member 4993]

(x-2)⁴ =0

will have

4 roots - (2, 2, 2, 2) - but all those are repeating roots. The graph of the function will be really (really , really , really ) flat at x = 2.

 

[Qwertyuiop[]]

How many roots did you find for x^10 - 8(x^8) + 16(x^6)= 0 ?

3. x=0 and x= +-2

 

[Deleted member 4993]

3. x=0 and x= +-2

You found 3 unique roots - and in total 10 roots including repeating roots.

 

[Steven G]

First (x⁵−4x³)²=0 is a 10th degree equation, not a 5th degree equation.

I like to think through what I do.
I would first ask myself what do I square to get 0. The answer is just 0
So (x⁵−4x³)=0
Like you, I would factor next, getting x³(x²−4)=0
Then x³=0 or (x²−4)=0
Giving us the solution x=0, x=2 or x=-2
Notice that there is no need to be concerned with taking the square root.