Theory of cubic equation question

[lionely]

Homework Statement

Find the equation who roots are the squares of the roots of the equation x^3-4x^2+x-



is the answer


x^3-16y^2+y+1?

 

[SammyS]

Homework Statement

Find the equation who roots are the squares of the roots of the equation x^3-4x^2+x-

is the answer

x^3-16y^2+y+1?

Neither of these is an equation.

An equation has an equal sign with a mathematical expression on each side of the equal sign..

 

[lionely]

omg i didn't notice my keyboard has a problem ......

it's x^3-4x^2+x-1=0

and my answer was
x^3-16x^2+x+1=0

 

[Michael Redei]

That second expression isn't an equation. But have you actually worked out what steps you need to take to solve the problem? Here's a hint:

Equations can have roots.
Roots have squares.
These squares can be roots of other equations.

 

[lionely]

I know i took the roots of the 1st equation to be x so the ones for the new equation are x^2.

I said let y= new roots, y= x^2
x= sqrt y

and I put x in first equation and I got x^3-16x^2+x+1=0

 

[Dick]

I know i took the roots of the 1st equation to be x so the ones for the new equation are x^2.

I said let y= new roots, y= x^2
x= sqrt y

and I put x in first equation and I got x^3-16x^2+x+1=0

Can you show all of what you did? I think that's way oversimplified. This is a problem involving symmetric polynomials, isn't it?

 

[lionely]

x^3-4x^2+x-1=0

new roots = x ^2
let y=x^2, x = sqrt y

(sqrty)^3-4(sqrty)^2+sqrty - 1
(y^3/2 - 4y + sqrty -1)=0
squaring both sides
y^3+16y^2+y+1

 

[Dick]

x^3-4x^2+x-1=0

new roots = x ^2
let y=x^2, x = sqrt y

(sqrty)^3-4(sqrty)^2+sqrty - 1
(y^3/2 - 4y + sqrty -1)=0
squaring both sides
y^3+16y^2+y+1

No, that is not the square. That's just bad algebra. E.g. (sqrty-1)^2 isn't y+1.

 

[lionely]

D:, I'll leave it as (y^3/2 - 4y + sqrty -1)=0 then.

 

[Dick]

D:, I'll leave it as (y^3/2 - 4y + sqrty -1)=0 then.

I think you missed the lecture were they were actually talking about how to do this problem. Sure, you could do that, but it's not a polynomial and it's not particularly well defined either if all of the roots aren't positive. What you missed is that if your polynomial has three roots a,b,c then you can write it as (x-a)(x-b)(x-c) then the coefficients of your original polynomial are various symmetric polynomials of a,b and c. The polynomial you want is (x-a^2)(x-b^2)(x-c^2). The coefficients are also symmetric polynomials of a, b, and c. Doesn't that ring some kind of bell?

 

[lionely]

oh.. Oh Oh Let me do it over

 

[lionely]

Is it a^6+3 a^4 y+7 a^4+3 a^2 y^2+14 a^2 y-5 a^2+y^3+7 y^2-5 y-1?

 

[Dick]

Is it a^6+3 a^4 y+7 a^4+3 a^2 y^2+14 a^2 y-5 a^2+y^3+7 y^2-5 y-1?

Not likely. Why do you even think it would be that? Could you start giving reasons why you think the answers you post are true?

 

[lionely]

Isn't it

y= x-a^2

so x = y+a^2? then i put it into the equation?

 

[Dick]

Isn't it

y= x-a^2

so x = y+a^2? then i put it into the equation?

That makes no sense. Expand (x-a)(x-b)(x-c)=0 and compare it with x^3-4x^2+x-1=0. You can express the numbers in the second polynomial as functions of a,b,c. What are they? You really haven't talked about this kind of thing in class or in the textbook?

 

[SammyS]

@lionely,

Look at the following post from Dick again !

I think you missed the lecture were they were actually talking about how to do this problem. Sure, you could do that, but it's not a polynomial and it's not particularly well defined either if all of the roots aren't positive. What you missed is that if your polynomial has three roots a,b,c then you can write it as (x-a)(x-b)(x-c) then the coefficients of your original polynomial are various symmetric polynomials of a,b and c. The polynomial you want is (x-a^2)(x-b^2)(x-c^2). The coefficients are also symmetric polynomials of a, b, and c. Doesn't that ring some kind of bell?

Multiplying (x-a)(x-b)(x-c) out gives: \displaystyle x^3 - (a+b+c)x^2+(ab+bc+ca)x-abc\ .

Equate this and \displaystyle x^3-4x^2+x-1\ .

You will not be able to solve for the a, b , or c individually, but you can find \displaystyle (a+b+c)\,,\ (ab+bc+ca)\,,\ \ and \ \ \displaystyle (abc)\ \ by inspection.

Now multiply out (x-a²)(x-b²)(x-c²) . With a bit of algebra and some cleverness, you will be able find your desired polynomial.

 

[micromass]

lionely, here is a LaTeX guide on how to post mathematical equations: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
It would help us a lot if you made your math symbols more readable.

 

[lionely]

Is it √x(x+1) = 1 + 4x ?

 

[Dick]

Is it √x(x+1) = 1 + 4x ?

No. Go back and read the previous posts and think about them. If you don't want to do that, then forget this problem and move on to the next one.

 

[Curious3141]

I know i took the roots of the 1st equation to be x so the ones for the new equation are x^2.

I said let y= new roots, y= x^2
x= sqrt y

and I put x in first equation and I got x^3-16x^2+x+1=0

Your method is basically sound, but I'm betting you screwed up the algebra. I did it your way and got the right answer. You have to be very careful in grouping terms and squaring out the surd forms.

While others have suggested using a slightly more involved method to compare the sums/products of roots to the coefficients, this is unnecessary here.

 

[lionely]

I'm trying, it's just because my teacher said do it this way, "Let y= to the new roots" and then put it into the first equation and then you will get your new equation...

 

[Curious3141]

x^3-4x^2+x-1=0

new roots = x ^2
let y=x^2, x = sqrt y

(sqrty)^3-4(sqrty)^2+sqrty - 1
(y^3/2 - 4y + sqrty -1)=0

Right so far.

squaring both sides
y^3+16y^2+y+1

Here is where you went wrong. That is NOT the correct square!

Firstly, express y^{\frac{3}{2}} as y\sqrt{y}.

GROUP all the terms with \sqrt{y} on the LHS, then group the other terms with integral powers of y (including the constant term) on the RHS. THEN, square both sides, simplify. Remember there are TWO terms on each side of the equation.

Remember: {(a+b)}^2 = a^2 + 2ab + b^2!

 

[lionely]

so is it ... {√x(x+1)}² = 1 + 16x²?

 

[Curious3141]

so is it ... {√x(x+1)}² = 1 + 16x²?

What?! How are you getting that? I just told you your working was right to a certain point and told you how to proceed from there. Why don't you just try that?

AGAIN, remember: {(a+b)}^2 = a^2 + 2ab + b^2!

 

[Dick]

Your method is basically sound, but I'm betting you screwed up the algebra. I did it your way and got the right answer. You have to be very careful in grouping terms and squaring out the surd forms.

While others have suggested using a slightly more involved method to compare the sums/products of roots to the coefficients, this is unnecessary here.

You are right. There is an easier approach. Nice.

 

[lionely]

SIGH , okay (y^3/2 - 4y + √y -1)

= y.√y -4y + √y -1

= y.√y + √y = 1+ 4y
= √y(y+1) = 1 + 4y

then i square both sides?

so i get

y(y+1)² = 1 + 16y^2

 

[Curious3141]

You are right. There is an easier approach. Nice.

Thanks, yes, this is an elementary approach, but the OP is messing up the algebra and failing to see why. It's the simple binomial square that he's getting wrong.

 

[lionely]

sighh sigh sighj sigh sigh sigh sigh why didn't i see that ...

(1+4x)^2 = 1 + 8x + 16x^2...

 

[Curious3141]

SIGH , okay (y^3/2 - 4y + √y -1)

= y.√y -4y + √y -1

= y.√y + √y = 1+ 4y
= √y(y+1) = 1 + 4y

then i square both sides?

so i get

y(y+1)² = 1 + 16y^2

(1 + 4y)^2 \neq 1 + 16y^2

In general, {(a+b)}^2 \neq a^2 + b^2.

Do you get that?!

 

[Dick]

Thanks, yes, this is an elementary approach, but the OP is messing up the algebra and failing to see why. It's the simple binomial square that he's getting wrong.

Yeah, that's been a problem all along. I'll leave the algebra correction up to you. I think lionely just did it again.

 

[Curious3141]

sighh sigh sighj sigh sigh sigh sigh why didn't i see that ...

(1+4x)^2 = 1 + 8x + 16x^2...

VERY GOOD. Finally. OK, now just simplify and group the terms.

Remember to expand the LHS correctly too.

BTW, if you're working in the variable 'y', you should use that throughout, otherwise it's confusing (and actually wrong).

 

[lionely]

it is x³ -14x²+x-1

 

[Curious3141]

it is x³ -14x²+x-1

Not quite. Every coefficient is right except the one for the x term. You must have made a mistake, show your exact work.

A bit of a cheat, but you can test your answer here: http://www.gyplan.com/eqcubic_en.html

And, not to belabour the point, what you wrote is still not an equation! You MUST put the '=0'.

 

[lionely]

lol... x^3 -14x^2-7x-1

 

[Curious3141]

lol... x^3 -14x^2-7x-1

Still not an equation. Just an expression. But all the coefficients are right.

 

[lionely]

looooooool OKAY HERE IT IS x^3 -14x^2-7x-1 = 0!

 

[Curious3141]

looooooool OKAY HERE IT IS x^3 -14x^2-7x-1 = 0!

GREAT!