# Homework Help: Quadratic equation with roots of α and β

1. May 15, 2010

### Maatttt0

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

N/a

3. 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

2. May 15, 2010

### phyzguy

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

3. May 15, 2010

### HallsofIvy

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

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.

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 $\alpha^2+ \beta^2= 11$.

Last edited by a moderator: May 15, 2010
4. May 15, 2010

### Maatttt0

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

5. May 15, 2010

### 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.

6. May 15, 2010

### Maatttt0

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

Thank you for the input Mentallic :)

7. May 15, 2010

### Mentallic

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

8. May 15, 2010

### 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

9. May 15, 2010

### Maatttt0

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

10. May 15, 2010

### Mentallic

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

Keep this in mind.

Now after cubing we have,

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

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,

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

Expanding,

$$-15u^{2/3}+75u^{1/3}-105=0$$

Substitute this into the other equality,

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

What do you end up with?

11. May 15, 2010

### Maatttt0

Sorry- still not getting it :|

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

12. May 15, 2010

### Mentallic

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

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

What we have is,

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

which is equivalent to,

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

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.

13. May 15, 2010

### Maatttt0

I fully understand now - thank you very much :D

14. May 19, 2010

### Unit

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