Solving Polynomial Roots: Sum of Cubes and Fourth Powers - Further Maths Help

[rdajunior95]

Show that the sum of the cubes of the root of the equation

x³ + (lambda)x + 1 = 0

is -3

Show also that there is no real value of (lambda) for which the sum of the fourth powers of the roots is negative.

This question is in one of the past papers of further maths and I don't know how to start solving this problem!

So please help me :)

P.S. Sorry if this is the wrong section, I am new here. ;)

 

[Office_Shredder]

If m is a root of the equation, what do you know is true about m³?

 

[Mark44]

Show that the sum of the cubes of the root of the equation

x³ + (lambda)x + 1 = 0

is -3

Your problem statement threw me off for a minute. I'm sure you mean "the sum of the cubes of the roots of the equation.."

Let r₁, r₂, and r₃ be the three roots of this equation.

1) What does it mean for r₁ to be a root of that equation?
2) Same for r₂.
3) Same for r₃.
4) What do you get if you add together r₁³ + r₂³ + r₃³?

Show also that there is no real value of (lambda) for which the sum of the fourth powers of the roots is negative.

This question is in one of the past papers of further maths and I don't know how to start solving this problem!

So please help me :)

P.S. Sorry if this is the wrong section, I am new here. ;)

 

[rdajunior95]

Your problem statement threw me off for a minute. I'm sure you mean "the sum of the cubes of the roots of the equation.."

Let r₁, r₂, and r₃ be the three roots of this equation.

1) What does it mean for r₁ to be a root of that equation?
2) Same for r₂.
3) Same for r₃.
4) What do you get if you add together r₁³ + r₂³ + r₃³?

Yes we have to show that r₁³ + r₂³ + r₃³ is -3

 

[Mark44]

What about questions 1, 2, and 3?

 

[rdajunior95]

when you equate x=r1, the equation would be equal to 0

 

[Mark44]

when you equate x=r1, the equation would be equal to 0

That makes no sense. An equation can't be equal to something. An equation is true or it isn't true, one or the other.

What does it mean to say that r₁ is a root of the equation?

 

[rdajunior95]

Yeah like when x=r1, r2 or r3 than the equation is true!

Sorry about the confusion!

 

[Mark44]

Which equations are you talking about? (These are what I'm trying to get you to show me, without much success so far.)

 

[rdajunior95]

(x^3) + (lambda)x + 1 = 0 this is the equation which is satisfied when x=r1, r2 or r3We have to show that (r1)^3+(r2)^3+(r3)^3 = -3

 

[Mark44]

What equation do you get when x = r₁?
What equation do you get when x = r₂?
What equation do you get when x = r₃?

Instead of just telling me about it, please write the equations.

 

[rdajunior95]

I think you have to use some kind of formula like when a cubic equation has roots (alpha), (beta) and (gamma) than to find the sum of α2 + β2 + γ2 you basically use

α² + β² + γ² = (α + β + γ)² - 2(αβ - βγ - αγ)

 

[rdajunior95]

I hope this helps! :)

 

[HallsofIvy]

Helps? This was your question and you haven't done what was suggested yet.

What does it mean to say that $x= r_1$ is a root of the equation $x^3+ \lambda x+ 1= 0$? And, as Mark44 said, "Instead of just telling me about it, please write the equations."

 

[Mark44]

I think you have to use some kind of formula like when a cubic equation has roots (alpha), (beta) and (gamma) than to find the sum of α2 + β2 + γ2 you basically use

α² + β² + γ² = (α + β + γ)² - 2(αβ - βγ - αγ)

I don't think that this will get you anywhere, even after fixing the sign error. The identity is α² + β² + γ² = (α + β + γ)² - 2(αβ + βγ + αγ).

If you can show that the sum of the three roots is 0, it's simple to show that the sum of the cubes of the roots is -3. Are you sure that what you have posted is the exact wording of the problem?