# Easy polynomials question?

If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2 -6x + a$ then find the value of "$a$" if $3\times \alpha + 2\times \beta = 20$
Thank you for the help

There is also a second question of this sort, i dont get that either. Would help if both were answered.

if $\alpha$ and $\beta$ are zeroes of the polynomial $x^2 -5x + k$ such that alpha- beta = 1. Find K

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How is the 2nd requirement related to the first? The first is a function of x, the second is function of alpha and beta? Also please add "homework" tag if its. –  Emmad Kareem Sep 1 '12 at 10:23
I seriously dont understand how to do it either. Also it isnt homework, well least this wasnt given to me in school, im skipping ahead. –  Aayush Agrawal Sep 1 '12 at 10:25
At least get the variables corrected. Alpha is not 'a' and beta is undefined. –  Emmad Kareem Sep 1 '12 at 10:28

First of all don't confuse $a$ with $\alpha$ !(It is better to substitute A or m for $a$)
We know that the sum of the roots of a quadratic equation $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$
So: $\alpha +\beta =6$
On the other hand:$3\times \alpha + 2\times \beta =20$
Solving this two equations two variables yields:$\beta = -2$ and $\alpha=8$
We also know that the product of the roots of a quadratic equation $ax^2+bx+c$ is : $x_1 \times x_2=\frac{c}{a}$
so $$\alpha \times \beta = -16 = a$$

Second question:
Again:
We know that the sum of the roots of a quadratic equation $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$
So: $\alpha +\beta =5$
On the other hand:$\alpha - \beta =1$
Solving this two equations two variables yields:$\beta = 2$ and $\alpha=3$
We also know that the product of the roots of a quadratic equation $ax^2+bx+c$ is : $x_1 \times x_2=\frac{c}{a}$
so $$\alpha \times \beta = 6 = k$$

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Where does alpha+beta=6 come from? –  Emmad Kareem Sep 1 '12 at 10:30
Thanks alot! I LOVE YOU! Also emmad, as stated -b/a = the sum. But a = 1 and b = -6, so -(-6) = 6. –  Aayush Agrawal Sep 1 '12 at 10:33
@EmmadKareem: As It is mentioned in the answer:We know that the sum of the root of a quadratic equation $ax^2+bx+c$ is $$x_1+x_2=-\frac{b}{a}$$ And for proving this, you should use the general quadratic formula(If you meant this) –  Zeta.Investigator Sep 1 '12 at 10:35
In the second question, we have $\alpha+\beta=5$ and $\alpha-\beta=1$. Note the general identity $$(\alpha+\beta)^2-(\alpha-\beta)^2=4\alpha\beta,\tag{1}$$ which can be easily verified by expanding the squares. Putting $\alpha+\beta=5$ and $\alpha-\beta=1$ we get $4\alpha\beta=24$, so $\alpha\beta=6$.
Remark: The solution procedure by PooyaM is better, since it works in both questions. However, the identity $(1)$ is sometimes useful, so I thought I would mention it.