Quadratic equation with roots of α and β

[Maatttt0]

Homework Statement

The quadratric equation x^2 - 5x + 7 = 0 has the roots α and β. Find the quadratic equation with roots α^3 and β^3.

Homework Equations

N/a

The Attempt at a Solution

u = α^3
α = u^(1/3)

u^(2/3) - 5(1/3) + 7 = 0

u^(1/3)[u - 5] = -7 --- minus 7 off each side followed by taking cubic root u out of the left side.

u[u - 5] = -343 --- put everything to the power of 3.

u^2 - 5u + 343 = 0

but the answer is u^2 – 20u + 343 = 0

I cannot see where I've gone wrong - help me please :) thank you

 

[phyzguy]

Well, for starters, when you cubed both sides of the equation, you forgot to cube the (u-5) term...

 

[HallsofIvy]

Homework Statement

The quadratric equation x^2 - 5x + 7 = 0 has the roots α and β. Find the quadratic equation with roots α^3 and β^3.

Homework Equations

N/a

The Attempt at a Solution

u = α^3
α = u^(1/3)

Surely you mean u= x^3 and x= u^(1/3)

u^(2/3) - 5(1/3) + 7 = 0

u^(1/3)[u - 5] = -7 --- minus 7 off each side followed by taking cubic root u out of the left side.

u[u - 5] = -343 --- put everything to the power of 3.

but you didn't. You should have u[u- 5]^3= -343- and multiplying that out gives a 4th degree equation not a quadratic equation.

u^2 - 5u + 343 = 0

but the answer is u^2 – 20u + 343 = 0

I cannot see where I've gone wrong - help me please :) thank you

where you've gone wrong is in not doing what you thought you did! You did not cube the entire equation.

You should try, instead, (x- \alpha^3)(x- \beta^3)= x^2- (\alpha^3+ \beta^3)x+ \alpha^3\beta^3.

knowing that \alpha and \beta satisfy x^2- 5x+ 7= 0] tells you that \alpha+ \beta= -5 and \alpha\beta= 7.

That makes it easy to see that \alpha^3\beta^3= (\alpha\beta)^3= 7^3.

For \alpha^3+\beta^3 use the fact that \alpha^3+ \beta^3= (\alpha+ \beta)(\alpha^2+ \alpha\beta+ \beta^2).

Note that since \alpha+ \beta= -5, (\alpha+ \beta)^2= \alpha^2+ 2\alpha\beta+ \beta^2= \alpha^2+ 2(7)+ \beta^2= 25 so that \alpha^2+ \beta^2= 11.

 

[Maatttt0]

Thank you for the replies guys - I'll have another go :)

 

[Mentallic]

or another way of doing it:

u^{2/3}-5u^{1/3}+7=0

u^{1/3}(u^{1/3}-5)=-7

on cubing both sides, u(u-15u^{2/3}+75u^{1/3}-125)=-343

But we know from the first equation that u^{2/3}-5u^{1/3}+7=0 so then -15(u^{2/3}-5u^{1/3}+7)=-15u^{2/3}+75u^{1/3}-105=0

Subbing this into our result, u(u-20)=-343 hence u^2-20u+343=0 as required.

 

[Maatttt0]

Ahh - this is what I was attempting to do but failed :P

Thank you for the input Mentallic :)

 

[Mentallic]

I had a feeling that that was the approach you were looking for

 

[Maatttt0]

I see when I was taking u^(1/3) out I only left u inside instead of u^(1/3) which messed the rest of the answer up :P

 

[Maatttt0]

But we know from the first equation that u^{2/3}-5u^{1/3}+7=0 so then -15(u^{2/3}-5u^{1/3}+7)=-15u^{2/3}+75u^{1/3}-105=0

Subbing this into our result, u(u-20)=-343 hence u^2-20u+343=0 as required.

Sorry - how did you get the u(u-20) part. Slightly confused by that part :S

 

[Mentallic]

<br /> u^{2/3}-5u^{1/3}+7=0<br />

Keep this in mind.

Now after cubing we have,

<br /> u(u-15u^{2/3}+75u^{1/3}-125)=-343<br />

But we want to get rid of the fractional powers, and notice how the u^{2/3} term has a multiplier of -15, let's multiply the first equation by -15.

So looking at,

<br /> -15(u^{2/3}-5u^{1/3}+7)=0<br />

Expanding,

<br /> -15u^{2/3}+75u^{1/3}-105=0<br />

Substitute this into the other equality,

<br /> u(u-15u^{2/3}+75u^{1/3}-125)=-343<br />

What do you end up with?

 

[Maatttt0]

Sorry- still not getting it :|

from u(u-15u^{2/3}+75u^{1/3}-125)=-343 to.. u(u - 20)

 

[Mentallic]

Ok how about this,

let y=u^{2/3}-5u^{1/3}+7

therefore -15y=-15u^{2/3}+75u^{1/3}-105

What we have is,

<br /> <br /> u(u-15u^{2/3}+75u^{1/3}-125)=-343<br /> <br />

which is equivalent to,

<br /> <br /> u(u-15u^{2/3}+75u^{1/3}-105-20)=-343<br /> <br />

Notice that part in there that is equal to -15y. Let's substitute that into it then,

u(u-15y-20)=-343

Do you get this?

But the only difference is that we had y=0, so you get the answer that you need.

If you don't understand the process, don't use it. HallsofIvy's method is perfectly fine too.

 

[Maatttt0]

I fully understand now - thank you very much :D

 

[Unit]

For \alpha^3+\beta^3 use the fact that \alpha^3+ \beta^3= (\alpha+ \beta)(\alpha^2+ \alpha\beta+ \beta^2).

\alpha^3+ \beta^3= (\alpha+ \beta)(\alpha^2 - \alpha\beta+ \beta^2).