Homework Help: Roots of a quadratic equation.

1. Dec 30, 2011

Darth Frodo

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

$\alpha$ and $\alpha^{2}$ are two roots of the equation x$^{2}$ -12x + k = 0

Find 2 values for k.

3. The attempt at a solution

$\alpha$ + $\alpha$$^{2}$ = 12
$\alpha$$^{3}$ = k

I have no idea where to go from here. Any help appreciated.

Last edited by a moderator: Dec 30, 2011
2. Dec 30, 2011

Staff: Mentor

Solve for α in the equation α2 + α - 12 = 0. Then substitute into your second equation to find k.

3. Dec 30, 2011

Darth Frodo

a(1 + a) = 12
a = 12 a = 11

12$^{2}$ - 12(12) + k = 0
k = 0

11$^{2}$ -12(11) + k = 0
k = 11

Is this right, because the answers at the back of my book are k = -64 k = 27

4. Dec 30, 2011

Staff: Mentor

This is not how to solve a quadratic equation. Read what I said in post #2.

Last edited: Dec 30, 2011
5. Dec 30, 2011

Darth Frodo

I apologize both to you and myself for such an error in my judgement.

Is there a rule as to when you are allowed to factorize via taking out what's common? Does the equation have to equal zero?

6. Dec 30, 2011

Staff: Mentor

Let me be a bit pedantic here by saying that an equation does not equal a number. An expression can be equal to a number, but an equation already has an '=' mark in it, so it would be meaningless to say that an equation is equal to zero or any other number.

What you might be alluding to is a theorem that says if the product of two numbers is 0, then one or the other of the numbers, or both, must be 0.

What you had was a(a + 1) = 12, from which you concluded that a = 12 or a = 11. This doesn't work at all. There are lots of ways that two numbers can multiply to 12, such as 1*12, 2*6, 3*4, (1/3)*36, -1*(-12), etc., etc. If you had checked your work, by replacing a with 12, you would have seen that 12(13) ≠ 12. Similarly, if a = 11, you would have seen that 11(12) ≠ 12.

But, by writing the equation as a2 + 1 - 12 = 0, and then factoring or using the Quadratic Formula, you should come out with the values that are solutions to the equation.

7. Dec 30, 2011

Darth Frodo

Ah ok, I see. Thanks for the clarification!